LIBRARY OF CONGRESS. 

i -.yA5*H^itt# T)c 






UNITED STATES OF AMERICA. 



THE ELEMENTS 



SPHERICAL TRIGONOMETRY. 



BY 



EUGENE L. KICHAEDS, B.A., 

ASSISTANT PBOFESSOS OF MATHEMATICS IX TALE COLLEGE, 



S>~ 



H 



IOC 



tfEW YORK: 
D. APPLETON AXD COMPANY, 

54a AND 551 BROADWAY. 

1879. 






COPYRIGHT BY 

B. APPLETON & COMPANY, 

1&79. 






PEEFAOE. 



In the following pages the author has endeavored 
to make the subject of Spherical Trigonometry plain 
to beginners. In order to do this, considerable space 
has been devoted to the treatment of the right-angled 
spherical triangle, and to the consideration of the ob- 
lique-angled spherical triangle, by means of diagrams, 
before applying the principles of trigonometry to each. 
It is believed that in this way the student will become 
familiar with the magnitudes about which his calcula- 
tions are made, and thus will be unwilling to use sym- 
bols and numbers in formulas without having clear 
ideas of what these symbols and numbers represent. 

In teaching some classes, it would not be necessary 
to go through the whole book. For instance, the solu- 
tion of right-angled triangles can be taught without 
using the chapter on that subject, by applying Napier's 
Rule of the Circular Parts to each example. The 



iv PREFACE. 

chapter, however, is convenient for reference in cases 
of difficulty. 

As the work was designed to be bound with " The 
Elements of Plane Trigonometry," the articles are num- 
bered as continuing that book. The references ac- 
cordingly are to " Plane Trigonometry," to Todhunter's 
" Euclid," and to Chauvenet's " Geometry." 



CONTENTS. 



CHAPTER XL 

PAGE 

Definitions. — Theorems of Right-angled Triangles, . . . 113 

CHAPTER XII. 
Solution of Right-angled Triangles, 127 

CHAPTER XIII. 
Quadrantal Triangles, 138 

CHAPTER XIY. 

Theorems and Formulas of Oblique-angled Triangles, . . . 143 

CHAPTER XV. 
Solution of Oblique-angled Triangles, 185 



CHAPTER XI. 

DEFINITIONS. THEOREMS OF RIGHT-ANGLED TRIANGLES. 

Art. 101. Spherical Trigonometry teaches the 
methods of finding the unknown parts of triedral an- 
gles from certain known parts, by means of the solution 
of spherical triangles. 

102. A spherical triangle is formed by the inter- 
section of the planes of a triedral angle with the surface 
of the sphere, the vertex of the triedral angle being at 
the centre of the. sphere. The sides of the spherical 
triangle are arcs of great circles (Ch. Art. 26, VIII.), 
which measure the face angles of the triedral angle 
(Ch. Art. 53, II.). The angles of the spherical triangle 
are equal to the diedral angles of the triedral angle 
(Ch. 16, YIIL). 

103. The face angles of a triedral angle are assumed 
to be each less than two right angles. Therefore each 
side of a spherical triangle will always be assumed as 
less than 180°. Each angle of a spherical triangle is 
less than 180°. 

(a) The trigonometric functions of a side of a tri- 
angle are the trigonometric functions of the angle 
measured by the side. 

101. A right triedral angle is a triedral angle one 
of whose diedral angles is a right diedral angle (Ch. 
Art. 43, YL). 



114 THE ELEMENTS OF SPHERICAL TRIGONOMETRY. 

A right-angled spherical triangle is a spherical tri- 
angle one of whose angles is a right angle. It is formed 
on the surface of a sphere by the planes of a right trie- 
dral angle, whose vertex is at the centre of the sphere. 

The hypotenuse is the side opposite the right angle. 

105. A quadrantal triangle is a spherical triangle 
one of whose sides is a quadrant. It is formed on the 
surface of a sphere by the planes of a triedral angle 
(whose vertex is at the centre of the sphere), one of 
whose face angles is a right angle. 

106. In a right-angled spherical triangle, not a 
quadrantal triangle, the three sides are each less than 
90° ; or, of the three sides, one is less than 90°, and the 
other two are each greater than 90°. 



Let OAJSChe a right triedral angle, of which each 
of the face angles is less than 90°. The planes of these 
face angles will form on the surface of the sphere a 
triangle ABC, each of whose sides is less than 90° 
(Art. 102). Let the triangle be represented on the sur- 



THEOREMS OF RIGHT-ANGLED TRIANGLES. 115 

face of a hemisphere of which the base is the circle 
BCB'O. Let the planes of the face angles, AOC srnd 
A OB, produced intersect the surface of this hemi- 



sphere in the arcs CA C , BAB' (which are semicircles 
(Ch. Art. 32, VIII.), and let the face BOO coincide 
with the base of the hemisphere. 

By hypothesis the triangle ABC has its three sides 
each less than 90°. 

In the triangle BAG', BA is less than 90° ; BO f 
and A C are each the supplement of an arc less than 
90°, and are, therefore, each greater than 90°. 

In the triangle B'A C,AC\s less than 90° ; B 'G and 
B'A are each greater than 90°, being each the supple- 
ment of an arc less than 90°. B'OO = BOC\ there- 
fore the arc B'G — the arc BO, and is less than 90°. 
Consequently in the triangle B ! AG, one side, B r C, is 
less than 90°, while the other two sides, B'A and O A, 
are each greater than 90°, being each the supplement 
of an arc which is less than 90°. 



116 THE ELEMENTS OF SPHERICAL TRIGONOMETRY. 

The triangles ABC, BAC, CAB', and B'AC are 
the only, four kinds of triangles, not quadrantal, into 
which the surface of the hemisphere can be divided. 
The four triangles, on the surface of the other hemi- 
sphere, formed by producing the planes AOC and 
A OB to meet the surface of that hemisphere, will be 
symmetrical to these (Ch. Art. 63, VIII.), and will 
therefore have their sides either each less than 90°, or 
one side less than 90°, and the other two sides greater 
than 90°. 

The principle stated therefore holds true for the 
whole surface of the sphere. 

107. Two angles or two arcs, or an angle and an 
arc, are said to be of the same species when both are 
less than 90°, or both are greater than 90°. 

108. In a right -angled spherical triangle, a side 
about the right angle and its opposite angle are always 
of the same species. 

Let a triangle ABC be constructed on the surface 
of a hemisphere having its sides each less than 90°, and 
let the figure be completed as in Art. 106. Then the 
hypotenuse will be greater than either of the other sides. 

For from C draw, in the plane BOC,& straight line, 
CD, perpendicular to the radius OB. CD will be 
perpendicular to the plane AOB (Ch. 18, VI.), and 
therefore perpendicular to the straight line AD, in that 
plane, drawn from D to A (Oh. Art. 6, VI.). 

Now OD + DA> OA (Euc. 20, I. Ch. 17, I.) ; 
therefore OD + DA > OB, and taking away. OD 

DA>DB; 
therefore the chord CA is greater than the chord CB 
(Euc. 47, I. Ch. 4, VI.), and the arc CA is greater than 



THEOREMS OF RIGHT-ANGLED TRIANGLES. 117 

the arc CB. Since the arc CA is greater than the arc 
CBj the angle CB A is greater than the angle GAB 
(Oh. 26, VIII.) ; but CB A is an angle of 90° ; there- 



fore the angle CAB is less than 90° ; but the side CB 
is by hypothesis less than 90° ; consequently the side 
CB and its opposite angle are of the same species. 

In a similar manner it can be proved that AB and 
the angle A CB are of the same species. 

In the triangle BAG, BA and BCA are both less 
than 90° ; BO is the supplement of BC, which is less 
than 90°, and is therefore greater than 90° ; but its op- 
posite angle BA C f is also greater than 90° being the 
supplement of BAC, which has just been proved to be 
less than 90°. 

In the triangle BAG, the two sides BA, B'C, and 
their opposite angles B'CA, B'ACnre all greater than 
90°, being the supplements of parts of the triangle 
BAG, which are all less than 90°. 

In the triangle B'AO, B' O and its opposite angle 



llg THE ELEMENTS OF SPHEEICAL TKIGOXOMETRY. 

A are both less than 90°, while B A and the opposite 
angle RCA are both greater than 9 

The principle of the article holds true for the trian- 
gles on the surface of the hemisphere represented by 
the figure ; :: ;->; holds true for the triangles syinmetri- 
_ - :; these on the mrfaae :: the : :her hemisphere, and 
therefore hoi Is true I : r rhe surface of the whole sphere. 
O the triangle had two right angles, the sides oppo- 
site these — ould be quadrants Oh. Art. S9. VIIIj. but 
the remaining side and its opposite angle would be of 
it species Ch. 16, VJJLL . 
109. In a right-angle ' ncal triangle, the s-ine 

of either of ~ rides about the right angle is equal 
to the product of the sine of the hypotenuse by the 
' -_/-" : V.~ 

Let ABC be a right-angled 
triangle on the surface of a sphere 
whose senb a h "he vert-::. '_'. :: 
the triedral angle, which forms 
the triangle by the intersection 
of its planes with the surface. 
Let B be a right angle, then 
OB is the edge of a right diedral angle (Art. 104). 
Ilien we are to pr:~e sin. a = an. : mt, A. 
From any poiie - F in OA itsw FI> in the plane 
AOCj snAFE in the plane AOB : t ::h straight lines 
- ::_/„: :_'_e- : OA, Let FD meet OC at Z>, and let 
FEmsxA ?JB - I. Draw the stnrighl tine 2>jE 

0A is perpendicular to the plan a Zi /CJB Qh •" • T 'I*)- 
The plane OAB is perpendicular to the plane DFE 

* This theorem can be easily remembered by its resemblance to (1) 




THEOREMS OF RIGHT-ANGLED TRIANGLES. 119 

(Ch. 17, VI.). The plane OBC is perpendicular to the 
plane OAB by hypothesis. 

Thus two planes, DFE, and OBC, are perpendicu- 
lar to a third plane OAB ; therefore their line of inter- 
section DE is also perpendicular to the plane OAB 
(Ch. 20, VI.). 

Consequently DEE, and DEO, are right-angled tri- 
angles (Ch. Art. 6, VI.). 

DEO and EFO are right-an- °^x^— ^_ c 
gled triangles by construction. \>^Ve /] 

DEE is the £>Z<m£ <m<7^ of *\ s --... / \ 
the diedral angle whose edge is \ V^ / 

OA, and therefore is equal to jl ^~ ^ ~b 

the angle A of the spherical tri- 
angle ABC (Ch. Arts. 39 and 45, VI. Prop. 16, 
VIH.). 

Denote the sides opposite the angles of the spherical triangle by 
small letters of the same name. 

sin. a = sin. BOG {(a) Art. 103) = — '((1) Art. 4) 





DF DE. 




DO DF' 


T)JP 

but = sin. CO A = sin. h ; 

DO 




T)Tp 

and — sin. DEE = sin. J. ; 





. • . (1) sin. a = sin. h sin. JL 

By drawing ED, and i£Z) perpendicular to OCj it 
can be proved that 

(2) sin. c = sin. b sin. (7. 

110. In the preceding article the demonstration has 
been applied to a figure drawn to represent a triangle 



120 TBLE ELEMENTS OF SPHEEIGAL TRIGOXOMETKY. 

of which the sides are each less than 90°, but the theo- 
rem is true for all right-angled triangle-. 

Let ABC represent a triangle on the surface of a 




hemisphere. Let the three sides, a, 6, and c, be each 
lesE than 9 ,: . Then the angles, ^4. and C are each less 
than 90° 'Art. 108). Complete the figure as in Art. 106. 

Not considering the right angle, of the five parts 
(three sides and two angles) of each of the triangles 
BAC\ B AC\ and B AC, one part, as in the triangle 
BAC, is the same as a part of ABC; or two part- 
in the triangle BA C, are equal to parts in the triangle 
AB C\ and the other parts are the supplements of parts 
of the triangle ABC. Therefore the theorem is true 
for these triangles (Art. 46, {a) Art 1 3). 

Thus in the triangle BAG, 
:::. B'C=im. a: sin. AC= rin. b; sin. CAB' 
= sin. A : but, according to previous article, 
sin. a = sin. b sin. A : substituting, 
sin. £ C = sin. AC X sin. CAB'. 



THEOREMS OF RIGHT-ANGLED TRIANGLES. 



121 



In a similar manner the theorem may be proved true 
for the triangles BA C and B'A C. 

If A and B were both right angles the theorem 




would still be true, for in that case AC and BG would 
be quadrants (Ch. Art. 89, VIIL)> and sin. a, sin. S, and 
sin. A would each be 1 ; also c would equal G (Ch. 16, 
VIII.). 

111. In a right-angled spherical triangle, the cosine 
of the hypotenuse is equal to the product of the cosines 
of the other two sides. 

Ox 

Let ABC 'be a triangle right- 
angled at _Z?, and on the surface 
of a sphere whose centre is 0, 
the vertex of the right triedral 
angle OABG Then we are to 
prove that cos. h = cos. a cos. c. 

Construct the triangle DEF as in the figure of 
Art. 109. 




122 THE ELEMENTS OF SPHERICAL TRIGONOMETRY. 

cos. I = cos. AOC = — ((4) Art. 4) 

OE OF 

x 



OD OE' 
Of? OE 

but — — - = cos. BOO — cos. a\ and = cos. AOB 

OJJ OE 

= cos. o ; 
therefore cos. b = cos. a cos. c. 

This proposition may be proved to be true for all right-angled trian- 
gles by applying the principle of Art. 46, with regard to the cosine, to 
the figure of the preceding article. 

112. In a right-angled spherical triangle the tan- 
gent of either side about the right angle is equal to the 
product of the tangent of the opposite angle by the sine 
of the other side* 

Let ABC be a spherical tri- 
y-^"/^ — -^ c angle, right-angled at B, formed 
\J^Qe /[ by a right triedral angle whose 

■^\ •--... / J vertex is at the centre, 0, of the 
\ .X / sphere. 

jj<^L— — ~Jb Then we are to prove 

tan. a = tan. A sin. c. 
Construct the triangle DEF %& in Art. 109. 
Then DEO, and EEO, are triangles right-angled 
at F\ and DEE, and DEO, are triangles right-angled 

at E. (See Art. 109.) 

tan. a = tan. BOC=—((2) Art. 4) ; 
EO vw ; 

^i= ^X |^= tan. DFEX sin. AOB 
EO EF EO 

= tan. A sin. c ; therefore 

* This theorem may be remembered by its resemblance to (2) Art. 32. 



THEOREMS OF RIGHT-ANGLED TRIANGLES. 123 

(1) tan. a — tan. A sin. c. 

By drawing FD and ED at right angles to OC it 
may be proved 

(2) tan. c = tan. (7 sin. a. 

113. In a right-angled spherical triangle, the tan- 
gent of either side about the right angle is equal to the 
product of the cosine of the adjacent oblique angle by 
the tangent of the hypotenuse* 

Construct the figure as in Art. 109. Then we are 
to prove tan c = cos. A tan. J. 

a no EF EF FD 

tan. c = tan. AOB = — — = — - x — — ; 

FO FD FO ' 

EF 
but ==- = cos. DFE = cos. A ; 
FD ' 

FD 

and = tan. A G — tan. b ; 

FO 

. • . (1) tan. c = cos. A tan. b. 

In a similar manner it may be proved 

(2) tan. a = cos. C tan. b. 

The theorem of this article and the theorem of the preceding article 
can both be shown to be true for all right-angled triangles, by applying 
Art. 46 to the figure of Art. 110. 

114. The Circular Parts of a right-angled spherical 
triangle are the sides ahoitt the right angle, the comple- 
ments of the hypotenuse and the oblique angles. 

The right angle is not considered one of the circu- 
lar parts. 

Of the five circular parts any one may be taken as 
the middle part, and then the two next to this, one on 
the one hand and the other on the other (not counting 

* This theorem may be remembered by its resemblance to (2) Art. 29, 




124 THE ELEMENTS OF SPHERICAL TRIGONOMETRY. 

the right angle as a part) are called adjacent parts, and 
the remaining two, each separated from the middle 

part by an adjacent part, are called 

opposite parts. 

Thus a, c, 90°-6, 90 9 -^4, and 90°- C, 
are the circular parts of the triangle ABC 
right-angled at B. 

If a is the middle part, 
adjacent parts are c, 90°— C; opposite parts 
are 90°-^, 90°-6. 

If c is the middle part, 
adjacent parts are a, 90°— A ; opposite parts are 90° — 6, 90°— C 

If 90° — b is the middle part, 
adjacent parts are 90°— J., 90°— C; opposite parts are a, c. 

If 90° — J. is the middle part, 
adjacent parts are 90° — 6, c ; opposite parts are a, 90°— C. 

If 90°— C is the middle part, 
adjacent parts are a, 90° — 6; opposite parts are c, 90°— A 

115. Napier } s rule of the Circular Parts. 

The sine of the middle part is equal to the product 
of the tangents of the adjacent parts ; and the sine of 
the middle part is equal to the product of the cosines 
of the opposite parts. 

Let ABC be a spherical triangle right-angled at B. 

We shall take each circular part in succession as a middle part ; be- 
ginning with 90°— 6, and next taking 90°— C> and so on. 

Taking 90°— b as the middle part we are to prove 
sin. (90°-5) = tan. (90°- A) tan. (90°- (7); or (Art. 5), 
(1) cos. b = cot. A cot. C. 
Now, according to Art. 112, 

(a) tan. a = tan. A sin. c, and 

(b) tan. c — tan. Csin. a. 



THEOREMS OF RIGHT-ANGLED TRIANGLES. 



125 



Divide equation (b) by equation (a), putting the 
second member of (b) over the first member of (a), 
tan. G sin. a tan. c 



tan. a 
tan. G cos. a = 



tan. A sin. <? ' 



(c) cos. a cos. c — 



tan. J. cos. o 
1 



((&) Art. 64), or 



= cot. A cot. G 



tan. ^4 tan. (7 
((f) Art. 64) ; but, according to Art. Ill, 
cos. b — cos. a cos. c ; therefore from (<?) 
cos. b = cot. .4 cot. C. Again, 
sin. (90°— b) = cos. a cos. c, or 

(2) cos. 5 = cos. a cos. <? (proved 
under Art. 111). 

Next take 90°— G as middle 
part, and we are to prove 

sin. (90°— G) = tan. a tan. (90°— I), or 

(3) cos. G — tan. a cot. J, and 

sin. (90° -67) = cos. c cos. (90°- A), or 

(4) cos. (7= cos. sin. ^L. 

To prove (3) we have from Art. 113 
cos. G tan. b = tan. a ; therefore 




„ tan. & 

cos. 6 = 

tan. 5 

To prove (4) 



tan. a cot. J ((/) Art. 64). 



/ 7\ /7 7 sin. a cos. J 
(a) cos. G — tan. a cot. b = X - 7 



cos. a 



sin. b 



((b), (cl), Art. 64); 
sin. & 



sin. J. 



sm. 



(Art. 109). 



126 THE ELEMENTS OF SPHERICAL TRIGONOMETRY. 




Also cos. h = cos. a cos. c (2) ; 
substituting these values for 
their equivalents in second mein- 
a ber of (d) 

cos. C = cos. c sin. A. 
Let a be the middle part, then 
we are to prove 

(5) sin. a = tan. c tan. (90° — C) = tan. c cot. C; 

(6) sin. a = cos. (90°— 5) cos. (90°— J.) 

= sin. h sin. ^1. 
To prove (5) 

sin. a = ^^ (Art. 112) 
tan. C 
= tan. c cot. £7 ; 
(6) is proved under Art. 109. 
Let c be the middle part, then 

(7) sin. e = tan. <z cot. A ; proved in the same 
manner as (5). 

(8) sin. c — sin. h sin. C (Art. 109). 
Lastly, let 90° — A be the middle part, then 

(9) cos. A = cot. 5 tan. c ; proved by Art. 113 ; see 
proof of (3). 

(10) cos. A = cos. a sin. C ; 



(e) cos. J. 
of Art. 61), 
sin. C 



cos^ x E*if ((9), and (S) and (3) 

sin. 5 cos. 



sin. £ 



sin. J 

tuting these values in (e) 
cos. A = cos. a sin. (7. 



; also cos. S = cos. a cos. <? ; substi- 



CHAPTER XII. 

SOLUTION OF RIGHT-ANGLED TRIANGLES. 

Art. 116. Any two parts (in addition to the right 
angle) of a right-angled spherical triangle being known, 
the triangle can be solved, and the unknown parts can 
be obtained in terms of the known parts, by applying 
Napier's rule. 

117. To find any unknown part from two known 
parts, care should be taken in applying Napier's rule to 
make such a selection of one of the three parts for a 
middle part that the other two should be either both 
adjacent parts or both opposite parts. 

118. Care should also be taken to observe the signs 
of the trigonometric functions, as the sign of the result 
will determine generally whether the arc or angle is 
< 90° or > 90° (Art. 46). 

119. The two parts, which may be given (in ad- 
dition to the right angle) to solve a right-angled spher- 
ical triangle may be : 1. the hypotenuse and a side ; 2. 
the hypotenuse and an angle; 3. the two sides about 
the right angle ■ 4. a side and an adjacent angle ; 5. 
a side about the right angle and the opposite angle / 
6. the two angles. 

120. Case I. — The hypotenuse and a side, of a 



128 THE ELEMENTS OF SPHERICAL TRIGONOMETRY. 



right-angled spherical triangle, being known, to solve 

the triangle. 

In the triangle ABC, right-angled at B, suppose 

b and a to be known. It is required to find c and the 

angles A and C in terms of a and b. 

To find c take 90°— b as the 

middle part, and a and c as the 

opposite parts ; then 

cos. b = cos. a cos. c (Art. 115) ; 

, v cos. b 

(a) or cos. c — . 

cos. a 

To find A take a as a middle 
part, 90°— JL and 90°— & as ^> 
posite parts ; then 

sin. a = sin. A sin. 5 ; whence 

, Z \ - a sin. a 

(6) sin. JL = . 

sin. 5 

To find C take 90°— 6' as a middle part, « and 
90° — 5 as adjacent parts ; then 
(c) cos. C = tan. # cot. b. 
As from equation (a) c is obtained by means of its 

cosine, the sign of — - — will determine whether c is 
cos. a 

< 90° or > 90° (Art. 46). If a and b are both < 90° 




or both > 90°, 



cos* b 



cos. a 



will be positive, and cos. c will 



also be positive, and therefore c will be < 90°. If a 
and b are not of the same species (Art. 107), — '- — will 



cos. a 



be negative, and consequently cos. c will be negative, 
and, therefore, c will be > 90°. 




SOLUTION OF RIGHT-ANGLED TRIANGLES. 129 

Though from (ft) A is obtained by means of its sine, 
A can have only one value, since A and a are of the 
same species (Art. 108). 

Also as G is found from its cosine, the sign of 
tan. a cot. ft will determine whether C is < 90° or 
> 90°. Again C and c must be of the same species 
(Art. 108). 

As check on work, form an 
equation in which only the three 
required parts occur. In the pres- 
ent case after c y A, and G are 
found, if the results are correct, 
the equation made by applying 
Napier's rule to 90°— C as a mid- A 
die part, c and 90°— A as opposite parts, that is 

cos. G = cos. c sin. A, 
should be a true equation. 

If it prove not to be a true equation, there must be 
some error in the previous calculations. 

121. Case II. — The hypotenuse and an angle of a 
right-angled spherical triangle fteing known, to solve 
the triangle. 

B being the right angle, suppose ft and A are given 
to find C, a, and c. 

Take 90°— ft as a middle part, 90°— A and 90°— G 
as adjacent parts ; then 

eos. ft = cot. A cot. G (Art. 115) ; 

e x . n COS. ft 

(a) cot. G— . 

cot. A 

To find a, use it as a middle part, and 90°—^., 

90°— ft as opposite parts. 

(ft) sin. a = sin. ft sin, A. 



130 THE ELEMENTS OF SPHERICAL TRIGONOMETRY. 



To find Cj take 90°— A as middle part, and c, 
90°— b as opposite parts. 

cos. A = tan. c cot. b, whence 
cos. A 



(c) tan. c 



cot. b 




As (7 is found by means of its cotangent, and as c 
is found by means of its tangent, 
the sign of the equivalent frac- 
tion in each case will determine 
whether the quantity is < 90° or 
> 90° (Art. 46). 

Though a is found by means 
of its sine, it can have but one 
value, being < 90° or > 90°, ac- 
cording as A is < 90° or > 90° (Art. 108). 
As check on work, use the equation 
sin. a — cot. C tan. c. 
122. Case III. — The two sides about the right angle 
of a right-angled spherical triangle being ~known, to 
solve the triangle. 

Suppose a and <?, about the right angle J3, to be 
known, it is required to find A, J, and C. 

To find A take c as a middle part, and 90°— A 
and a as adjacent parts ; then, 

sin. o — cot. A tan. a (Art. 115) ; 

(a) cot. A = . 

tan. a 

To find b y take it as middle part, 

(b) cos. b = cos. a cos. c. 

To find C take a as a middle part, and c, 90°— C 
as adjacent parts. 



SOLUTION OF RIGHT-ANGLED TRIANGLES. 



131 



sin. a 



(c) cot. C-- 



cot. C tan. c y or 
sin. a 



tan. <? 




As A and Care each found by means of its cotan- 
gent, and as b is found by means of its cosine, each of 
these quantities can have only one 
value, which will be determined 
by the sign of its trigonometric 
function (Art. 46). 

As check on work, use the 
equation 

cos. b — cot. A cot. C. 

123. Case IV. — A side about 
the right angle, of a right-angled spherical triangle, and 
the adjacent oblique angle being known, to solve the tri- 
angle. 

B being the right angle, suppose c and A are known ; 
it is required to find 5, C, and a. 

To find 5, take 90°— A as the middle part, and 

90°— by 0, as adjacent parts. 

cos. A = cot. b tan. c (Art. 115), or 

, . ,7 cos. A 

{a) cot. 6 = . 

tan. c 

To find Cy take 90°— (7 as middle part, and (?, 

90°— A as opposite parts. 

(5) cos. (7 = cos. sin. A. 

To find #, take c as middle part, and #, 90° — ^L as 

adjacent parts. 

sin. ^ = tan. a cot. J., or 

sin. c 



(c) tan. # = 



cot. A 



132 THE ELEMENTS OP SPHERICAL TRIGONOMETRY. 




As the required parts are found by means of the 
cotangent, cosine, and tangent respectively, each can 
have but one value, which will be determined by the 
sign of its trigonometric function (Art. 46). 
As check on work, use the equation 
cos. G = cot. b tan. a. 

124. Case Y. — A side about 
the right angle, of a right-angled 
spherical triangle, and the oppo- 
site angle being known, to solve 
the triangle. 

B being the right angle, a and 
A are given ; it is required to find 
by C y and c. 

To find b, take a as a middle part, and 90°— A, 
90°— b as opposite parts. 

sin. a == sin. b sin. A (Art. 115); 

, . . -, sin. a 

(a) sin. o = . 

sin. A 

To find Cy take 90°— A as middle part, and a, 
90°— C as opposite parts. 

cos. A == cos. a sin. (7; whence 

/7A • n cos. .4 

(b) sin. C— ■. 

cos. a 

To find Cy make it the middle part, and #, 90°— .A 
adjacent parts. 

(<?) sin. c = tan. a cot. -4. 

As, in this case, each of the required parts is ob- 
tained by means of its sine, each part will have two 
values (Art. 46), and there will be solutions answering 
to two triangles, each of which will contain the given 
parts. 



SOLUTION OF RIGHT-ANGLED TRIANGLES. 133 

The required parts of one will be supplements of 
the required parts of the other. 

As check on work, use the equation 

sin. c = sin. h sin. C. 

125. That there will be two right-angled spherical 
triangles containing the same given parts, when those 
parts are a side and the opposite angle, will be evident 
from the accompanying figure. 




Let BA'OAC be an ungula in a hemisphere of 
which the base is the circle, APA', and the centre 
0. The lune ABA'C will be the base of the wedge 
(Ch. Art. 90, VIII.). 

Let a plane BOPO be passed through the pole, P, 
of the circle ABA', through the centre, 0, and any 
point, B, of the arc ABA' (Ch. Art. 33, VIII.). The 
weclge will be divided into two right triedral angles, 
OB AC, OBA'C, having the face angle, BOO, in com- 
mon, and having the diedral angles, A and OA', the 



134 THE ELEMENTS OF SPHERICAL TRIGONOMETRY. 

same, since these are parts of the diedral angle AA r . 
The remaining parts of the one are the supplements of 
the corresponding parts of the other. 

The lune will be divided into two right-angled tri- 
angles, corresponding to the right triedral angles, hav- 




ing the side a in common, and having the angle A 
equal to the angle A! (Ch. 16, VIII.), and having the 
remaining parts of the one the supplements of the cor- 
responding parts of the other. 

This case of the solution of right-angled spherical 
c triangles is called the ambiguous 
case. 

126. Case VI.— The two an- 
gles of a right-angled spherical 
triangle leing known, to solve the 
triangle. 

B being the right-angle, sup- 




SOLUTION OF RIGHT-ANGLED TRIANGLES. 



135 




pose A and C to be known ; it is required to find b, 

a, and c. 

To find by make 90°— b the middle part ; then 

c (a) cos. b = cot. A cot. C 

(Art, 115). 

To find a, make 90°— A the 

a middle part. 

cos. A = sin. (7 cos. #, or 

/t\ cos. -4. 

(5) cos. a = - — -• 
sm. 6 

To find e 5 make 90° — <7 the middle part 

cos. (7 = cos. sin. .4, or 

, N cos. G 

(c) COS. c = . 

sin. ,4 

As check on work, use the equation 

cos. b == cos. # cos. <?. 



Suppose A, B, C to be a right-angled 
spherical triangle of which the hypotenuse, 
£, is 100° and the side, a, is 60°. Required 
to find the other parts. 




cos. b = cos. a cos. <?. 




sin. a = sin. b sin A. 


cos. b 




sin. a sin. 60° . 


cos. a 


sin. b sin. 100° 


cos. 100° 


— quantity. 


A of same species as a. 


• cos. 60° ~ 


. ■ . A < 90°. 


.-. c> 90°. 






Log. eos. 100° 


= 9.239670 


Log. sin. 60°= 9.937531 


Log. cos. 60° 


= 9.698970 


Log. sin. 100°= 9.993351 



Log. cos. 69° 40' 40.7" 
.'. c=110° 19' 19.3" 



9.540700 Log. sin. 61° 34' 6| ff 9.944180 



136 THE ELEMENTS OF SPHERICAL TRIGONOMETRY. 

(Check.) 
cos. C = cos. c sin. A. 
cos. C = tan. a cot. b. Log. cos. c = 9,540700 

= tan. 60° cot. 100° = — quantity. Log. cos. A = 9.944180 

.-. tf> 90°. 9.484880 

Log. tan. 60°= 10.238561 Log. cos. C — 9.484880 

Log. cot. 100°=: 9.246319 

Log. cos. 72° 13' 2.27" 9.484880 
.*. 0=107° 46' 5W3". 

(Examples worked by six-place logarithm tables, and answers given 
to nearest tenth of second.) 

Solve a right-angled triangle when there are given 

Hypotenuse and a side. 

Example 1. Hypotenuse, 72° ; side, 13°. Ans. Side, 71° 30" 35" ; 

angles, 13° 40' 54" ; 85° 41' 53". 
2. Hypotenuse, 64° 40' ; side, 137° 50'. 

Hypotenuse and an angle. 

8. Hypotenuse, 73°; angle, 15°. Ans. Angle, 85° 31' 14"; 

sides, 14° 19' 48.7" ; 72° 26' 11.8". 

4. Hypotenuse, 92° ; angle, 95°. Ans. Angle, 68° 15' 10" ; 

sides, 95° 23' 5" ; 68° 9' 55.6". 

Two sides ahout the right angle. 

5. 63° and 42°. Ans. Hypotenuse, 70° 16' 57i" ; 

angles, 71° 10' 25f ; 45° 18' 2\ 

6. 16° and 116°. Ans. Hypotenuse, 114° 55' 20.5" ; 

angles, 17° 41' 40" ; 97° 39' 24.4". 

7. 100° and 135°. 



A side about the rigid angle and an adjacent angle. 

8. Side, 39° ; angle, 61°. Ans. Hypotenuse, 59° 5' 29.4" ; 

side, 4S° 37' 34.3" ; angle, 47° 10' 45.5". 



SOLUTION OF RIGHT-ANGLED TRIANGLES. 137 

9. Side, 100° 35' ; angle, 50° 2'. Arts. Hypotenuse, 96° 50' 37.6" ; 
side, 49° 32' 55.3" ; angle, 98° 5' 31.5". 

10. Side, 112° 4' ; angle, 100° 6'. 

A side about the right angle and an opposite angle. 

11. Side, 25° 16' ; angle, 36° 13'. 

A _ ' 46° 15' 15.5" ; 

^• Hy P° tenUSe '133°44'44.5"; 

40° 7' 40" ; 63° 8' 36.4". 

81(161 139° 52' 20"; anglG ' 116° 51' 26.6". 

12. Side, 114° 2' ; angle, 102° 15'. 

Ans. Hypotenuse, ^.^'SJ.j 

29° 8' 13.2" ; 31° 23' 52.6". 

Slde ' 150° 51' 46.8" ; ang 6 ' 148° 36' 7.4*. 

13. Side, 62° 10' ; angle, 74° 1'. 

Two angles, 

14. 56° and 40°. Ans, Hypotenuse, 36° 30' 3f ; 

sides, 29° 32' 49.2" ; 22° 28' 45.6". 

15. 131° 20' and 110° 15'. Ans. Hypotenuse, 71° 3' 57.6" ; 

sides, 134° 44' 41" ; 117° 26' 53.9". 

16. 50° and 120°. 



CHAPTEE Xni. 



QUADRAJSTAL TRIANGLES. 



Art. 127. A quaclrantal triangle is formed by a trie- 
dral angle (at the centre of a sphere), one of whose face 
angles is a right angle (Art. 105). 

Thus suppose ABC is a triedral angle (at the cen- 
tre, 0, of a sphere), having one of its face angles, A C, 




a right angle. Then the planes of the faces of the trie- 
dral angle will form, by their intersection with the sur- 
face of the sphere, a quadrantal triangle, ABC, of which 
the side, AC, which measures the right angle, AOC, is 
a quadrant (Ch. Art. 53, II.). 



QUADRANTAL TRIANGLES. 139 

128. The polar triangle of a quadrantal triangle is 
a right-angled triangle. 

Let ABC be a quadrantal triangle, having its side, 
AC ov b, a quadrant. Let A'B'C be the polar triangle 




of ABC (Ch. Art. 67, VIIL). Denote the sides of 
the triangle ABC by a, b, c, and the sides of A'B'C' by 
a', b\ and c\ respectively. 

Then A f B'C is a right-angled triangle, right-an- 
gled at B r . 

B'+.b = 180° (Ch. Arts. 69 and 70, VIIL) ; there- 
fore 

B' = 180°- b = 180°- 90° = 90°. 

129. Any two parts of a quadrantal triangle, in 
addition to the side which is a quadrant, being known, 
we can solve the polar right-angled triangle. 

For the two given parts of the quadrantal triangle 
are the supplements of two parts of the polar triangle 
(Ch. 18, VIIL). Two parts of the polar triangle, 
in addition to the right angle (Art. 128), will there- 
fore be known, and the polar triangle can be solved 
(Art. 166). 

Thus let ABC be a quadrantal triangle, having the side, 5, a quad- 
rant. "Lei A'B'C be the polar triangle of ABC, Denote the sides of 
the triangle ABC by a, 6, c, and the sides of A'B' C by a', b\ and c', 
respectively. 



140 THE ELEMENTS OF SPHERICAL TRIGONOMETRY. 

(1) a' = 180°— -4 ; (2) V = 180°— -5; (3) c' = 180°— C; also 
(4) ^' = 180° — a; (5) E = 180° — 6 = 90° ; (6) C = 180° — c 

(Oh. Art. 70, VIIL). 

From equations (1) to (6) inclusive, it is evident that if we know any 

two parts of ABC, besides b, we know two parts of A'B' C besides the 

right angle B'. 

130. The polar right-angled triangle being solved, 
and all its parts being known, all the parts of the 




quadrantal triangle will be known, as the sides of the 
one are the supplements of the angles of the other 
(Ch. 18, VIIL). 

131. To solve a quadrantal triangle we derive from 
Arts. 128, 129, and 130, the following rule : 

Take the supplements of the given parts of tlte quad- 
rantal triangle for the given parts of the polar right- 
angled triangle ; solve the polar triangle, and take the 
supplements of the parts found as the required parts 
of the quadrantal triangle. 

If in the quadrantal triangle we have given, in addition to the quad- 
rant, the other two sides, in the polar right-angled triangle we have given, 
besides the right angle, the two angles ((4) and (6), Art. 129). 

If in the quadrantal triangle we have given any two angles, in the 
polar right-angled triangle we have given two corresponding sides ((1), (2) 
and (3), Art. 129). 

If in the quadrantal triangle we have given, in addition to the quad- 
rant, a side and an angle, in the polar right-angled triangle we have given, 



QUADRANTAL TRIANGLES. 



141 



besides the right angle, an angle and a side ((4) and (1) or (4) and (2), 
etc., Art. 129). 

In any example all these relations will be best seen by drawing the 
quadrantal triangle and its polar triangle. 



Example. Solve a quadrantal 
triangle of which there are given, 
in addition to the side which is 
a quadrant, a side equal to 64°, 
and an angle between the side 
and quadrant equal to 120°. 

Let b = 90° ; a — 64° ; 
C= 120°. 

.-.& = 90°; A' = 116°; 
c' = 60°. 

Solution of A'B'C falls un- 
der Case IY. (Art. 123). 

cos. A' = cot. b' tan. c'. 

cos. A' cos. 116 
. • . cot. b' : 




= — quantity . • . b' > 90° 



tan. c' tan. 60 
Log. cos. 116° = 9.641842 
Log. tan. 60° = 10.238561 

Log. cot. 75° 47' 49.43" = 9.403281 

6' = 104° 12' 10.57". 

sin. c' = cot. A' tan. a' 



cos. C = sin. A' cos. c' 
= sin. 116° cos. 60° = + quantity. 
.-. <7'<90°. 
Log. sin. 116°^ 9.953660 
Log. cos. 60°= 9.698970 



, tan. a' = - 



sin. c' sin. 60° 



cot. A cot. 116° 
= — quantity. 
, '. a'>90°. ' 
Log. sin. 60° = 9.937531 
Log. cot. 116° = 9.688182 



Log. cos. 63° 17' 41.95" = 9.652630 Log. tan. 60° 36' 44.8" = 10.249349 

C- 63° 17' 41.95". a'= 119° 23' 15.2". 

{Check.) cos. C'= cot. b' tan. a'. b'= 104° 12' 10.57" 

Log. cot. b'— 9.403281 . • . B = 75° 47' 49.43". 

Log. tan. a'= 10.249349 C'= 63° 17' 41.95" . * . c = 116° 42' 18.05/ 
Log.cos.C"= 9.652630 «'= H»° 28' 15.2' .'.A= 60° 36' 44.8". 



142 THE ELEMENTS OF SPHERICAL TRIGONOMETRY. 

Solve a qnadrantal triangle when there are given, in addition to the 
=i ic ~hicL is 2. "j 'zi ii -ir. ~ : 

Example 1. The two angles adjacent to the quadrant, W and 104°. 
^ju. Angle = 86° 24 r 36.5"; sides = 7: ffi ^S T and 103° 32 _ 
2. The two odier sides, ?0° and 1 

Ams. An^es, 84° 24' 11.9'; 69 3 15 45.1' ; 105 3 59' 15.3". 
S. A side 50% angle between the side and quadrant 12 : . 



CHAPTER XIV. 

THEOREMS OF OBLIQUE-ANGLED TRIANGLES. 

Art. 132. If from a point in the surface of a hemi- 
sphere, which is not the pole of its base, arcs of great 
circles are drawn to the circumference of the great cir- 
cle, which is its base, of all these arcs the greatest is 
that which passes through the pole, the least is that 
which when produced passes through the pole, and of 




the others that which is nearer to the greatest is greater 
than the more remote y and from this point equal 
arcs of great circles can be drawn only in pairs. 

Thus, let the figure PBECGr represent a hemi- 
sphere having for its base the great circle BECG. 



144 THE ELEMENTS OF SPHERICAL TRIGONOMETRY. 

Let P be the pole of the base (Ch. Art. 27, YIIL). 
Suppose A to be any other point on the surface of the 
hemisphere. From A let the arcs APB, A C, AD, AE, 
AF, and AK, be drawn to the circumference, BECG. 




Then, of these arcs APB is the greatest, AC is the 
least, and AD, which is nearer to APB, is greater than 
AE, which is more remote ; A E than AF, etc. 

Let M be the centre of the sphere of which the fig- 
ure represents the hemisphere. Join PM\ then PM 
is perpendicular to the plane of the circle, BECG (Ch. 
Art. 27, YIIL), and the planes, BPCM and BFCM, 
are perpendicular to each other (Ch. 17, VI.) ; also the 
arcs BAC and BFC are perpendicular to each other 
(Ch. Art. 58, VIII.). BC is the diameter of the sphere 
common to the two semicircles BPCM and BFCM 
(Oh. Art. 32, VIIL). 

Now, let a perpendicular be let fall from A upon 
the plane of the circle BECG. It will lie in the plane 
BACK (Ch. Art. 51, VL), and at L will meet at right 
angles the line of intersection, BC, of the two planes 
BA CM and BFCM (Ch. Art. 6, VI.). 



THEOREMS OF OBLIQUE-ANGLED TRIANGLES. 145 

Draw the straight lines ZD, ZE, ZF, and ZK, in 
the plane BFGG; also draw the straight lines AB, 
AD, AE, AF, AK, and AG. 

Now, M is the centre of the circle BFGG (Ch. 
Art. 26, VIII.), and L is a point in the diameter which 

P 




is not the centre; therefore LB is the greatest line 
which can be drawn from L to the circumference 
BFGG, LG is the least, and ZD, which is nearer to 
ZD, is greater than ZE, Zi? than ZF, etc. (Euc. 7, III. 
Ch. IT and 24, I.). 

Therefore of the straight lines, drawn from the 
point A to the points B, D, E, F, K, and G, AB is 
the greatest, AG is the least, and AD is greater than 
AE, AEttxm AF(Evlc. 47, 1. Ch. 4, VI.)! 

Consequently, since in equal circles, or in the same 
circle, greater chords cut off greater arcs, the arcs being 
less than semicircumferences (Ch. 6, II.) of all the great 
circle arcs, drawn from A to points on the circum- 
ference BFGG, APB is the greatest, AG is the 
least, AD is greater than AE, AE than AF, and 
AF than AK, these arcs being arcs of equal circles 



146 THE ELEMENTS OF SPHERICAL TRIGONOMETRY. 

and all less than semicireumferences (Ch. Art. 29 and 
32,VIIL). 

Again, from the point L equal straight lines can 
be drawn to the circumference JBFCG only in pairs 
(Euc. 7, III. Ch. 20, I.). Consequently, equal straight 
lines, and equal arcs of great circles 2 can be drawn from 
A to the circumference, BFGG^ only in pairs (Euc. 
47, I. and 28, III. Ch. 5, VI. and 5, II.). 

133. Let every point on the surface of the hemi- 
sphere be projected upon the plane of its base (Ch. 




Art. 56, VI.) ; that is, let the hemisphere itself be pro- 
jected upon the plane of its base. 

Let the figure BECG represent the hemisphere thus 
projected, with the arcs also projected in the lines 
BBC, DAH, AK, and AK 

The arc BPC of the preceding figure will be pro- 
jected in the straight line BG. Therefore the straight 
line BPG will represent an arc on the surface of the 
hemisphere perpendicular to the arc BECG. 



THEOREMS OF OBLIQUE-ANGLED TRIANGLES. 14? 

By this method of projection ACK, ACE, A CD, 
and ABD, ABE, etc., are made to represent triangles 
right-angled at C and at B, respectively. 

134. If arcs of great circles are drawn to the cir- 
cumference of the base of a hemisphere, from a point 
on the surface not the pole of the base, of the two adja- 
cent angles made by these arcs with the circumference 




of the base, the one on the side toward the longer per- 
pendicular is an obtuse angle, and the other angle on 
the side toward the shorter perpendicular is acute. 

Let the hemisphere be represented as projected on 
the plane of its base, and let AB, AD, AE, AE, be 
arcs drawn from A on the surface of the hemisphere 
(not the pole of the base) to the circumference BECG. 
Of these arcs let BA C be the one passing through the 
pole of the base. BAG will be perpendicular to the 
arc BECG (Ch. Art. 58, YIII.). 

In the right-angled triangle ADB, AB is > AD 
(Art. 132); 



148 THE ELEMENTS OF SPHERICAL TRIGONOMETRY. 

. • . ABB is > ABB (Oh. 26, VIII.), i. e., is > 90°, 
or is an obtuse angle. 

Again, in the right-angled triangle ABC, AC is 
<AB (Art. 132); 

. • . the angle ABC is < the angle ACB (Ch. 26, 
VIII.), i. e., < 90°, or is an acute angle. 




In a similar manner it may be proved that the an- 
gles ABB, AKB, are obtuse, and that the angles 
ABC and AKC are acute. 

135. A perpendicular upon an arc, which coincides 
with a side of an oblique-angled spherical triangle, 
drawn from an angle opposite this side, falls within the 
triangle, if the other two angles are both acute, or are 
both obtuse / but falls without if one of these angles is 
acute while the other is obtuse. 

Let HAK, HAE, etc., be triangles represented on 
the surface of a hemisphere which is projected on the 
plane of its base (Art. 133). Let these triangles have 
the common vertex A, not the pole of the base. Let 



THEOREMS OF OBLIQUE-ANGLED TRIANGLES. 149 

P be the pole of the base. Let the great-circle arc, 
BAC, be drawn through jP, and let it meet the circum- 
ference of the base in two points, B and G The angles 
at B and G will be right angles (Ch. Art. 58, VIIL). 




Suppose the triangle HAK to have the angles H 
and K acute; also suppose the angles AEK, ADK, 
and A GH are acute. 

As the angle C is a right angle, A CM is greater 
than H; therefore AG is less than AH (Ch. 26, VIIL), 
and must be the shorter perpendicular (Art. 132), for 
in the same manner it may be proved less than AK or 
AE, or less than AD or AG; therefore it is the least 
arc from A, and it is perpendicular by hypothesis. 
Now, as A G is the shorter perpendicular, the acute an- 
gle H lies toward AG; for the same reason, the acute 
angle K lies toward AG (Art. 134) ; that is, G lies be- 
tween .ZTand K, or the shorter perpendicular, A C, falls 
within the triangle AKH. In the same manner it may 
be shown to fall within the triangle HAE. 



150 THE ELEMENTS OF SPHERICAL TRIGONOMETRY. 

A C having been shown to be the shorter perpen- 
dicular, AB must be the longer perpendicular. As by 
hypothesis ADE is acute, ADB must be obtuse, and 
must lie toward the longer perpendicular (Art. 134) ; 




and, since AGH is acute, AOB must be obtuse, and 
must lie toward the longer perpendicular; therefore 
AB lies within the triangle ADG, of which the angles 
AGB and ADG are obtuse. In the same manner it 
may be proved that AB lies within the triangle AEG, 
the angles A GE and AEG being obtuse. 

Next let ^.EZTbe a triangle having the angle AEK 
acute, but the angle AKE obtuse ; then both perpen- 
diculars, AC and AB, lie without the triangle. 

Since AEK\& acute, AC lies to the right of AE 
(Art. 134). Since AKE is obtuse, AKC must be acute, 
and AC lies to the right of AK; therefore AC, the 
shorter perpendicular, lies without the triangle. 

Again, since AKE is obtuse, .Allies to the left of 
AK. Since AEK is acute, AEB must be obtuse, and 



THEOREMS OF OBLIQUE-ANGLED TRIANGLES. 151 

AB lies to the left of AE\ therefore the longer per- 
pendicular also lies without the triangle. 

136. In a spherical triangle — 1. the sum of two an- 
gles is 180°, if the sum of the opposite sides is 180° ; 
2. is less than 180°, if the sum of the opposite sides is 
less than 180° ; 3. is greater than 180°, if the sum of 
the opposite sides is greater than 180°. 




Suppose ABC to be a spherical triangle projected 
on the plane of the base of the hemisphere ABDB'. 
Let DD be the projection of the arc drawn through 
6" perpendicular to the arc AB produced. 

1. Suppose (a-\-h) = 180°; then (A + .S) = 180°. 

If a — b, then a and h each = 90° ; and A and B 
each = 90° (Ch. Arts. 38 and 58, VIII.). 

.♦. (A + B = 180°. 

Let a and h be unequal. 

Then a and h must lie on the same side of DD '. 

For if i were on one side of DD' and a on the other 
side, occupying the positions GA and CB f respectively, 



152 THE ELEMENTS OF SPHERICAL TRIGONOMETRY. 

they would together form an arc, ACB', equal to a 
semicircle, and the figure ABB'G would be a lune and 
not a triangle (Ch. Arts. 32 and 90, VIII.). 

Therefore a and b both lie on the same side of DD'. 
Produce A C to meet the circumference of the base of 




the hemisphere in B'. Then ACB' is a semicircumfer- 
ence, and is equal to 180° (Ch. Art. 32, VIII.). 

Then b + CB' = 180° = b + a; 

.-. CB' = a=CB; 
and the angle CB'B = CBB' (Ch. 23, VIII.) = CAB 
(Ch. 16, VIII.); 

. • . CBB'= CAB ; add to each CBA ; 

CBB'-\- CBA = CAB + CBA = A + B; but 

CBB'-\- CBA = 180° ; 

therefore A + B = 180°. 

2. Suppose a + b < 180°; 
then A + B< 180°. 

Then C cannot be the pole of the arc ABE; for if 



THEOREMS OF OBLIQUE-ANGLED TRIANGLES. 153 

C were the pole, a and b would each be 90° (Ch. Art 37, 
VIII.) ; that is, a -\- b would equal 180°, which is con- 
trary to the hypothesis. 

If a — b, CD being the shorter perpendicular from 
C upon BB\ a and b are both greater than CD (Art. 




132) ; therefore B and B' are each < 90° (Art. 108), 
if CB' and CB are a and b ; therefore their sum is less 
than 180°. 

Suppose now a and b unequal, and a <b. Pro- 
duce A C to meet the circumf erence of the base of the 
hemisphere at E. From C draw the arc CE'— CE. 
Then b+OE = 180° (Ch. Art. 32, YIIL). 

.-.(tf + 5) <(&+<?.#); a< CE; 
also a< CE\ 

From C two arcs, each equal to a, may be drawn to 
EDE\ one above CD, between CD and CE, and the 
other below CD, between CD and CE (Art. 132), CD 



154 THE ELEMENTS OF SPHERICAL TRIGONOMETRY. 

being the shorter perpendicular* from G upon the cir- 
cumference of the base of the hemisphere. 




Suppose first that OB is taken as a. 

Now OB < CE, as just proved ; 

therefore the angle OEB < OBE (Ch. 26, VIII.). 

But OEB = CAB (Ch. 16, VIII.) ; 

therefore CAB < OBE; add OB A to each ; 

CAB + OB A are less than CBE+ CBA ; 



* CD is assumed as the shorter perpendicular ; but if proof that it is 
the shorter perpendicular were required, take 0, on the circumference of 
the base, as the pole of PCD and draw the arc CO. CO is an arc of 
90° (Ch. Art. 37, VIII.). C has been proved not to be the pole of the 
base ; therefore the proposition of Art. 132 applies, and the shorter per- 
pendicular lies to the right of CO. But again, a + b by hypothesis is 
less than 180°, and a < 6, therefore a is < 90° and < CO, and must lie 
to the right of CO. But the perpendicular is the least line from C to 
the circumference OE'BD, and is therefore less than a, or CB, and must 
lie to the right of a, or CB. 

If a is the perpendicular itself the proposition is still true, for it is 
then < 90° (Art. 108), and D = 90°. 



THEOEEMS OP OBLIQUE-ANGLED TRIANGLES. 155 

but CBE+ CBA = 180° ; 
therefore CAB + CBA < 180°, or 

A + B< 180°. 
Next suppose CB is taken as a. 
Since CD is the shorter perpendicular, 
CBA <90°; also CAB' < 90° (Art 134) ; 
therefore their sum CBA + GAB < 180°. 




3. Let a + J > 180° ; then also 

' A + ^ > 180°. 
Now, if (7 were the pole of AB, a and h would each 
equal 90° (Oh. Art. 37, VIII.), and a + h would equal 
*180°, which is contrary to the hypothesis ; therefore G 
is not the pole of AB. 

If a = J, a would lie on one side of CD' and b on 

the other side (Art. 132). Now, if we take O and O' 

on the circumference of the base, each at the distance 

of a quadrant from (7, since a and "b Qi a — b) are each 

3 



156 THE ELEMENTS OF SPHERICAL TRIGONOMETRY. 

greater than 90°, a and h must each meet the circum- 
ference OD'O' to the left of and 0' (Art. 132) ; there- 
fore in this case the angles A and B are each obtuse 
(Art. 134), and their sum is greater than 180°, since 
CD' is the longer perpendicular from C* 

If a and b are unequal, then one of them, as a, might 
be below CD\ and the other, as 5, might occupy one 




of two positions (Art. 132), as CA or CA', one below 
CD' and the other above CD'. 

If a be below CD', and I above CD', then A'CB 
will be the triangle to be considered. CA'B and 
CBA' are each obtuse in this case (Art. 134), and there- 
fore CA'B + CBA' > 180°. 



* This may be proved by a method similar to that employed in the 
note for 2 ; or A and B may be proved obtuse directly, thus : 

CD'> CB.' .B> D' (Ch. 26, VIII.) ; i. e., B > 90° ; in the same 
way the equal angle is > 90° ; . • . A + B > 180°. 



THEOREMS OF OBLIQUE-ANGLED TRIANGLES. 157 

If a and h are both below CD\ then CAB is the 
triangle to be considered. 

Produce AC and BC to meet the circumference of 
the base of the hemisphere again at H and K. Then 
ACK and BCK are semicircumferences (Ch. Art. 32, 
VIII.), and each equals 180°. 




. • . a+ CK= 180° and h+CH=: 180°, or 
a + b + £#+ <7iT= 360°, but a + 5 > 180°. 
.-. (75 r +C r X<180°; 
therefore by the 2d case 
(1) CHK+CKH< 180°, 
but (75^= <7AX= 1%0°-CAB = 180°-- J. ; 
also C2ff= CBff(Ch. 16, Vni.) == 180°-^. 
Substituting these values in (1) 
360°-(J. + 5)<180°; 
therefore A + B > 180°. 

137. The converse of the preceding article is true, 
that is, the sum of two sides of a spherical triangle is 



158 THE ELEMENTS OF SPHERICAL TRIGONOMETRY. 

180°, if the sum of the opposite angles is 180° ; is less 
than 180°, if the sum of the opposite angles is less than 
180° ; is greater than 1S0° 5 if the sum of the opposite 
angles is greater than 180°. 

Let two angles of a spherical triangle be denoted 
by A and B, and the sides opposite them by a and h 
respectively. 

1. If J. + .5 = 180°, a + 5 = 180°. 

Let A' and B' be the corresponding angles of the 
polar triangle, and a! and V the sides opposite these. 




Then A = 180°— a', B = 180°— J', 

A' = 180°-a, B ! £= 180°-J (Ch. 18, VIII.) ; 

. • . substituting for A and B their equivalents in 1, 

360°-<>' + V) = 180° ; or a' + V = 180° ; 

consequently A! + B' = 180° (Art. 136). ' 
Substituting values for A' and B\ 
360° -(a + h) = 180°, or a + h = 180°. 
2. If^ + ^<180°, + 5<18O°; 

for, passing to the polar triangle, 

360°-(a' + J')< 180°, or a' + V > 180°; 

consequently A! -f B' > 180° (Art. 136) ; that is, 
360°-(a -\-b)> 180°, or a + b < 180°. 



THEOREMS OF OBLIQUE-ANGLED TRIANGLES. 159 

3. If A + B > 180°, a + 1 > 180° ; 
for, passing to the polar triangle, 

360 o -(V + V) > 180°, or a' + b'< 180° ; 
consequently A' + B' < 180° (Art. 136) ; that is, 

360°-O + 1) < 180°, or a + I > 180°. 

138. The sines of the sides of a spherical triangle 
are proportional to the sines of the opposite angles. 





Thus let ABC be any spherical triangle, of which 
the angles are A, B, and (7, and the sides opposite 
them are a, 5, and <?, respectively. Then 

sin. a _ sin. A 

sin. h sin. B 

From C let the arc CD be drawn perpendicular 
to the side <?, or (in the right-hand figure) to c pro- 
duced, meeting it in the point D. Then ABC is equal 
to the sum or to the difference of two right-angled tri- 
angles. 

In the triangle A CD> taking CD as a middle part 
and 90°— 5, 90°— CAD as opposite parts, 

sin. CD = sin. I sin. CAD (Art. 115) 
= sin. h sin. A (Art. 46). 

In the triangle BCD, also taking CD as a middle 



part and 90 c 



-a, 90 c 



-B as opposite parts, 



160 THE ELEMENTS OF SPHERICAL TRIGONOMETRY. 

sin. CD = sin. a sin. B ; 

. • . sin. a sin. B = sin. b sin. A, or 

/ N sin. a sin. ^4 

<#> - — I = ~ »• 

sm. 6 sm. jtf 





In a similar manner it can be proved 
sin. a sin. A 
sin. C 



0) 



sm. c 

From (&) by alternation we have 
sin. ^L _ sin. B 
sin. # sin. b 
From (5) by alternation 
sin. ^1 _ sin. (7 
sin. a sin. c 
sin. A sin. J? 



(<0 



; therefore 
sin. G 



sm. <z 



sin. b 



sm. <? 



If the perpendicular coincided with the side a, B would be a right 
aDgle, and we should have from (a) 

sin. a = sin. 6 sin. A (Art. 35) 
an equation already established (Art. 109). 

139. Suppose a to be > b, then A is > B (Ch. 26, 
VIII.). 



THEOREMS OF OBLIQUE-ANGLED TRIANGLES. 161 



sin. a sin. A 



(Art. 138) ; 



(Ch. Art. 10, III.); 



sin. b sin. B 
sin. a -\- sin. b _ sin. A -f- sin. B 
sin. a— sin. b sin. A— sin. J? 
**-*(* + *) = t^HA±B) 
tan. i (a-5) tan.i (A-B) vv y ; 

140. If an arc of a great circle be drawn from one 
of the angles of a spherical triangle perpendicular to 
the opposite side, or to the opposite side produced, 
then 

(1) The sines of the segments of the side, on which 
the perpendicular falls, will be proportional to the nu- 
merical COTANGENTS of the ADJACENT ANGLES. 





Thus, in the triangle ABC, if the perpendicular 
CD is drawn from C to c, or c produced, 
sin. AD __ cot. A 
sin. BD cot. B 

For in the triangle A CD, taking AD as a middle 
part and 90°— CAD, CD, as adjacent parts, 

(a) sin. AD = cot. CAB tan. CD (Art. 115) ; 
and in triangle BCD, taking BD as a middle part and 
<7D, 90°— i?, as adjacent parts, 

(5) sin. ^2) = cot. B tan. CD ; 
dividing equation (a) by equation (S), 



162 THE ELEMENTS OF SPHERICAL TRIGONOMETRY. 



<*) 



sin. AD cot. CAB cot. A 



(Art. 46), 



sin. BD cot. 5 cot. B 
cot. -4 being the numerical value of cot. A. 

If the sign as well as the numerical value of cot. A 
be taken into account, equation (c) for the left-hand 





figure, or in the case in which the perpendicular falls 
within the triangle, would be unaltered ; but for the 
right-hand figure, or in the case in which the perpen- 
dicular falls without, (e) would become 
,y. sin. AD _ —cot. A 
sin. BD cot. B 

(2) The cosines of the segments of the side will he 
proportional to the cosines of the sides adjacent to 
them. 

In triangle A CD, taking 90°— b as a middle part, 

and CD y AD as opposite parts, 

cos. AD cos. CD — cos. b (Art. 115) ; 

and in triangle BCD, taking 90°— a as a middle part, 

and BD, CD as opposite parts, 

cos. BD cos. CD = cos. a ; 

dividing the first of these equations by the second, 

/ s cos. AD cos. b 

(e) = . 

cos. BD cos. a 



THEOREMS OF OBLIQUE-ANGLED TRIANGLES. 163 



141. Let ABC be a spherical triangle, and from 
C let an arc of a great circle, CD, be drawn perpen- 
dicular to the opposite side. 

Let the side a be > 5, then 
is A >B (Ch. 26, YIIL), De- 
note the segments BD and 
AD by x and y, respectively. 

1. Let the perpendicular 
fall within the triangle. 

Then, tan. ^(x— y) 




For sin * y = cotA 
sin. x cot. B 

tan. B 



sin. (A-B) , , 
sm. (J_ + ^ 

((1) Art. 140) 



tan. ^4 
sin, a? -f- sin, y _ tan. A -j- tan. B 
sin. a?— sin. y 



((f) Art. M); 

(Ch. Art. 10, III.) ; 



tan. A— tan. J? 
sin. A , sin. 5 
cos. J. cos. .Z? 
sin. A sin. .Z? 



((J) Art. 64); 



therefore 



tan, 



cos. A 



COS. i? 

sin.(^+^) 



tan. |- (x—y) sin. (JL 
(a) Art. 67, (a) Art. 68) ; or 
sin. {A-B) 



-B) 



((a) Art. 71, 



tan 



i (*-y) ■ 



sm.(A + £) 



tan. |- <?,* since x-\-y = c. 



*If a is >6 and a and 6 are each >90°, A and B are also >90° 
(Art. 136). Then y is >z (Art. 132), and the equation above will be 



tan. i (y— z) = 



sin. (^-^.) 
sin. {B+A) 



tan. £ c. 



164 THE ELEMENTS OF SPHERICAL TRIGONOMETRY. 

The last equation may be written 

t \ a i / \ sin. i- (A— B) cos. 4 (A— B) . 

(a) tan. 4 (x—y) = *-± { ^ 1 tan. 4 e 

sin. J (^1+JB) cos. ±(A+B) 2 

((a) Art. 73). 

2. Let the perpendicular fall 
without the triangle. 

BD and AD 5 denoted by x and 
y, are external segments. 




Then tan. ^ (a? -f- y) — 



For ^^ - ~ eot ^ 
sin. a? 

cot. CA2> 



cot. J? 



cot. i? 

(Art. 46) = 



sin. (J.-^) 
sin. (A + B) 

((d) Art. 140) 

tan. B 
t^n~CAB 



tan. 



((/)Art.64); 



sin. x + sip, y = tan. CAD + tan. J (Ch> ^ 
sin. #— sin. y tan. CAB— tan. i? 
sin. <7J.Z> , sin. ^ 



>os. (X4i) cos. B 
sin. C1P sin. B 
cos. 6L4Z> cos. _Z? 



((b) Art. .64); 



But .# is <^ (Ch. 26, VIII.), therefore B—A is a negative quantity, 
and sin. (£—A)=-sm. (A-B) (Art. 95). 

Also tan. J (y— x) = - tan. J (z— y) (Art. 95). 

Substituting these values our new equation becomes like the equa- 
tion above — 



sin. (A-B) 
tan. i (x-y) = . , . _~ ^n. ± , 

SID. (JL + 1?) 



THEOREMS OF OBLIQUE-ANGLED TRIANGLES. 165 



therefore taD " * ( *+^ = "M^P + f) ((g) Art. 71), 
tan. | (a?-y) sin. (CAD-B) vv 7 " 

(a) Art. 67, (a) Art. 68). 

Now, C-AZ? = 180°— 4, and substituting the value 

of CAD in the right-hand member of the last equation, 

tan. j (x+y) _ sin. {180°— (4-5)} 



tan. | (x— y) 



tan. $(x + y) 



sin. {180°-(4+5)} 
sin. (4—2?) 



sin. (A + B) 
sin. (4-5) 



, or since x—y—c, 



tan. ^ e. 



tan. £ c 



sin. {A+B) 
The last equation may be written 

(6) tan. ¥ (z+y) _ ^ ( ^_^ cog> ( ^ + ^ 

((«) Art. 73). 

142. From the angle, C, of a spherical triangle, let 
an arc of a great circle, CD, be drawn perpendicular to 
the opposite side or opposite side produced. 





y^T ^B 



Suppose a > h, then is A > B (Ch. 26, VIII.). 
Denote the segments of the base by x and y, re- 
spectively. 



166 THE ELEMENTS OF SPHERICAL TRIGONOMETRY. 



Then tan. %(a + b) tan. $ (a—b) 
cos. b _ cos. y 



tan. i (x+y) tan. \ (x—y). 



COS. & COS. X 



((2) Art. 140), 





cos. b — cos. a cos. y — cos. x tr ,, * , -, A ttt\ 

= 2 (Ch. Art, 10, III.), or 

cos. b -f- cos. a cos. y -\- cos. a? 

tan. I (a + b) tan. |- (a— b) — tan. |- (x + y) tan. |- (x—y) 
((b) Art. 71, (/) Art. 64). 

In the left-hand figure x + y = c ; 

in the right-hand figure x—y = £ ; therefore 

(«) when the perpendicular falls within the triangle 
tan. \(a-\-V) tan. -| (a—b) = tan. |- c tan. |- (x—y) ; * 

(5) when the perpendicular falls without the triangle, 
tan. \ (a + b) tan. |- (a—b) — tan. j- (a? + y) tan. J c. 



* When « is >b and a and 6 are each >90°, y is >z (Art. 132), and 
equation («) above would become 

tan. -£- (a + 5) tan. -§- (6 — a) = tan. -J- c tan. 4- (y—x). 
But -J (6 — a) is a negative quantity, 

tan. \ (b—a) = — tan. \ (a — b) (Art. 95) ; 
also tan. £ (y—x) = — tan. -J- (x—y). 
Substituting these values our new equation becomes like the equa- 
tion above — 

tan. \ (a 4- b) tan. J (a — 6) = tan. £ c tan. -J- (x—y). 



THEOREMS OF OBLIQUE-ANGLED TRIANGLES. 167 

143. In a spherical triangle, the tangent of half 
the sum of two sides equals the ratio of the cosine of 
half the difference to the cosine of half the sum of 
the opposite angles, multiplied by the tangent of half 





y^—-^i» 



the third side ; and the tangent of half the differ- 
ence of two sides equals the ratio of the sine of half 
the difference to the sine of half the sum of the op- 
posite angles, multiplied by the tangent of half the 

THIRD SIDE. 

Thus, a, b, and e denoting the sides of a spherical 
triangle, and A, B, and C denoting the angles respect- 
ively opposite these, a also being greater than b, 

(a) tan. A (a 4- b) = C0S ' *" ^ — — — - tan. i c. 
■ 2 ■ cos.±(^1 + jB) 



(5) tan. |- (a—b) 



sin. 4. (-4— J5) 



tan. £ 



sin.-i- (-4+^) 

1. When a perpendicular, drawn to the third side 
from the opposite angle, falls within the triangle, i. e., 
when A and B are both acute or both obtuse (Art. 
135). 

Denote the segments BD and AD by x and y, re- 
spectively. 



168 THE ELEMENTS OF SPHERICAL TRIGONOMETRY. 

tan.-|0-H) = t an. \ (A+B) (A ^ 
V ; tan. -J- (a-b) tan. I (A-B) V ' ; ' 



fc%\\ 1 / \ sin. 4 (J.— i?)cos.iM — i?) , ; 
(2) tan.fr(g— y)== -.— ?;. , ~ 77-j ^tan.^c 



((a) Art. 141). 



sin. %(A+B) cos.|- {A + B) 



tm^(a + b)t^.^fl-V) = ^ ^ ^ 
tan. \ (x-y) 
Multiply equations (1), (2) and (3) together, mem- 
ber by member, cancel like terms in numerator and 





denominator of the resulting fractions, and extract the 
square root of each member of the final equation, and 

(4) tan. i (a+ b) = '^TTzf! tan " * * 

COS. f (A+J3) 

Now equation (1) may be written 

tan. \ (a—b) _ tan. \ (A—B) 

tan. \ (a+b) ~~ tan. \ (A+B) 

Multiply the equation in this form by equation (4) and 



(5) tan. \ (a-b) 



.tin. J (A—B) 



tan. \ c. 



sin. £ (A+B) 

2. When the perpendicular falls without the trian- 
gle, i. e., when A is obtuse and B acute. 



THEOREMS OF OBLIQUE-ANGLED TRIANGLES. 169 

The proof is the same except that in equations (2) 
and (3) we have tan. ^ (x-\-y) in place of tan. \ (w—y) 
((b) Art. 141, and (b) Art. 142). 

144. In a spherical triangle, the tangent of half 
the sum of two angles equals the ratio of the cosine 
of half the difference to the cosine of half the sum 
of the opposite sides, multiplied by the cotangent of 
half the third angle; and the tangent of half the 
difference of two angles equals the ratio of the sine 
of half the difference to the sine of half the sum of 
the opposite sides, multiplied by the cotangent of half 

the THIRD ANGLE. 

Thus, in the triangle ABC, a being >b y 




(a) tm.%(A+£) 



(b) tan. I- (J. - 



-B) = 



cos. \ (a—b) 
cos. \ (a+b) 
sin. \ (a— b) 



cot. i- 0. 



sin, 



*(«+*) 



cot. A (7. 



Let J/.S'C" be the polar triangle of ABO. 
Since by hypothesis a is > 5, 
180°— a is <180°-J; bnt J/=180°— a, and 
^'=180°-5 (Ch. 18,YIIL); 



170 THE ELEMENTS OF SPHERICAL TRIGONOMETRY. 

therefore A' is <B '; consequently al is <V (Ch. 26, 
VIII.). 

(Art. 143) ; 

but a'=180°-J. ; b'=180°-A; c'=180°-C; 
^l'=180 o -a; B'=180°-b. 




Substituting these values in (1) 
tan. {180°-^ (A+B)} 



cos. \ {a— h) 
cos. U80°— $(a+J)} 



tan. {90°-$- 0\, or 



tan. \ (A+B) = cos - £<>-?> cot . £ tf (Art. 46, and 
cos. £ («+e>) 
Art. 5). 

Again, tan. J (&'-«') = sm ' \ \ ~ > tan. £ c 

sin. £(i?'+-4 / ) 

(Art. 143). 

Substituting for a', V, o', etc., tbeir values 



tan. \{A-B) = !^lfc|). cot. | a 
* V ; sin. *(«+*) 



THEOREMS OF OBLIQUE-ANGLED TRIANGLES. 171 

Equations (a) and (b) of this article, and equations (a) and (6) of the 
preceding article, written in the form of proportions, are called Napier^s 
Analogies. 

145. Tojmd an expression for the cosine of an an- 
gle in terms of the sides of a spherical triangle. 




Let ABC represent a triangle on the surface, cor- 
responding to the triedral angle, 0, at the centre of 
the sphere. 

From E 9 any point in the edge of the diedral angle 
OA, let DE be drawn in the plane OAC, perpen- 
dicular to OA, and let EF be drawn in the plane 
OAB, also perpendicular to A. Let ED meet 00 
in D, and let EF meet OB at F. Join D and F by 
the straight line DF. DF lies in the plane OBO 
(Euc. Def. 7, I. Ch. Del 6, Int.). 

DEE, the measure of the diedral angle whose edge 
is OA (Ch. Art. 39, and Art. 45, VL), is also the meas- 
ure of the angle A of the spherical triangle (Ch. 16, 
YIIL), and may be taken to represent that angle. 

«, 5, G y the sides of the spherical triangle, are the 
measures of the angles BOO, COA,AOB, respectively 
(Art. 102). 



172 THE ELEMENTS OF SPHERICAL TRIGONOMETRY. 



From triangle DEF 

H\ a nra . DE 2 +EF 2 -DF* 
(1) cos. A = cos. DEJ< = _ 

2DEEF 
((3) Art. 61). 

Since DEO and EOF are triangles right-angled at 




E, DE 2 =D0 2 -OE 2 , and EF 2 = OF 2 -OE 2 ; sub- 
stituting in (1) these values of DE 2 and EF 2 , 

. D0 2 +OF 2 -DF 2 -20E 2 

cos. A — ! . 

2DEEF 
Divide numerator and denominator of the fraction 
by 20D.OF, and 



cos. A = 



D0 2 +0F 2 -DF 2 
20D.OF 



OF OE 
OD X OF 



BE EF 
OD X OF 



^ D0 2 +OF 2 -DF 2 n/1B , 

JNow, ' ^^ ^,_, = cos. DOF-- 

' WD.OF 



cos. a 



((3) Art. 61) ; 
OE 



OE 



cos. DOE= cos. I ; — = cos. EOF— cos. c : 
OD ' OF 



THEOREMS OF OBLIQUE-ANGLED TRIANGLES. 173 



DE . , A EF . 

— = sin. b; and — =Bin.c; 

therefore 



cos. a — cos. b cos. c 
sin. b sin. 
In a similar manner it may be proved 
cos. b — cos. c cos. a 



(a) cos. A — 
In a similar ] 
(5) cos. ^ == 



sin. e sin. a 

, N ~ cos. tf — cos. a cos. 5 

(c) cos. 6 = : = — _ 

sm. a sin. 

146. To find an expression for the tangent of one 

c 




half an angle, in terms of the sides of a spherical 
tricmgle. 

A, J9, and C denoting the angles of the triangle, 
and a, b, e denoting the sides opposite these, s being 

a + b + c 

2 ' 

(a) tan. A A = /™;(^)™-(^ 

y sm. 5 sm. (s — a) 



(h) tan. I 2? = V siD - ('-°> Sin ; (g ~ a ) . 
y sm. 5 sin. (s—b) 



174 THE ELEMENTS OF SPHERICAL TRIGONOMETRY 

/ x . , n /sin. (s—a) sin. Cs—b) 

(c) tan. \0= J ? { v ; . 

y sin. 5 sin. (5— <?) 

For 

/ 7\ a cos - #— cos. b cos. , . l 

(d) cos. J. = ^ ^.^ (Art. 145). 




Subtract each member of the equation from 1„ 

-1 A sin. b sin. c + cos. b cos. c — cos. # 
1 — cos. A = ' 

sin. J sin. c 
= cos- (^-^)- cos. ^ ((5) Art 6g) 
sin. b sin. 

2 sm. ^ — = ) sm. ^ — ■ '- 

2 • 2 

sin. b sin. £ 
((d) Art. 70). 

Now, let, (l), g+ft+g = «; then, (2),^^= *-a; 

(3), -L_ = s-b; and, (4), ^— = s-c, 

substituting in the last equation the values of 

a-\-c— b 1 a + b—c 
— ! and — ■ . 

2 2 



THEOREMS OF OBLIQUE-ANGLED TRIANGLES. 175 

(e) 1-cos. A = 2 sin - <?-*> sin - ( s ~ c \ 
sin. sin. <? 

Add 1 to both members of equation (d\ then 

. . . cos. a — cos. J cos. c + sin. 5 sin. c 

1 + cos. ^1 = ; — . 

sm. 6 sm. 

_ cos. a— (cos. 5 cos. c — sin. & sin. c) # 
sin. 5 sin. 

. , . cos. a — cos. (&+c) //zx A nH , 

or 1+cos. ^ = : — - . v -^- y ((5) Art. 67) 

sm. sm. 

2 sm. ^ — ■ — — sm. ^— J 

2 ((<*) Art. 70); 



sin. 5 sin. 
substituting values of — and 



2 



, ,, N _, . A 2 sin. 5 sin. (s—a) 

(/) 1+cos. J. = . . v '-. 

sm. 6 sm. e 



Divide equation (e) by equation (f) and extract the 
square root of both members. 

/ 1-cos. A _ A* 
V 1+cos. A ~V 



-cos. ^1 _ /sin. (5— S) sin. (5— c) m 
1+cos. A V sin. s sin. (5— a) 



or, since tan. \ A = A / '- — ((c) Art. 75), 

V 1+cos. A 



r 



tan.iA = J ™ ; ('-S)™- ('-*). 
r sin. s sin. (s— <z) 

In a similar manner it can be proved 

tan. ±JB= A"- (»-*) «J°. («-q) . 

r sin. ^ sin. (5— &) 



176 THE ELEMENTS OF SPHERICAL TRIGONOMETRY 



tan. % C = 



sin. (s—a) sin. (s—b) 
sin. s sin. (s — c) 
147. To find an expression for the cosine of a side 
in terms of the angles of a spherical triangle. 

Let ABC he a spherical triangle. A, B, C being the 




angles, a, b, c being the sides opposite these. Let A'B' C f 
be the polar triangle of ABC, A' 9 J?', C being its angles 
and a\ b\ c f being the sides opposite these. 

(1) cos. A'= cos ' ^- cos - }' C0S - d ({a) Art. 145). 



^'=180°- 



-a\ 



sin. V sin. c' 
a'=180°-J.; V=1W°-B\ 



^ = 180°-(7(Ch. Arts. 69 and 70, VIII.) ; 
substituting these values in (1) 

-cos.^-cos.5cos.^7 (Art46)50r 



— cos. a : 



(a) cos. a = 



sin. B sin. C 
cos. ^1+cos. B cos. C 



sin. jB sin. CI 
In the same manner it may be proved 
cos. i?-}-cos. C cos. A m 



(b) cos. b = 



sin. (7 sin. J. 



THEOREMS OF OBLIQUE-ANGLED TRIANGLES. 177 

cos. C-\- cos. A cos. J? 



(c) COS. c 



sin. A sin. B 



148. To find an expression for the tangent of one 

HALF a SIDE, IN TERMS of THE ANGLES of a Spherical 

triangle. 

Let A, B, C denote the angles of the spherical trian- 
gle ; let a, b, c denote the sides opposite these respect- 
ively, and let S denote — - — -t— ; then 



cos. S cos. (S—A) 



, \ . -1 / — COS. /O cos. o — -ZLI 

(a) tan. i a = . / * ' 

V (cos. S- B) cos. [S- C) 

(b) tan. i I = A / -^S cob. (8-B) 

V cos. (S-C) cos. (£- A) 



(c) tan. i * = „-co 8 .gcoB.(g-P) > 

V cos. (S-^i) cos. (£-5) 

(d) For cos. a = C0S ' ^+ C08 ' * cos ' °. 

sin. i? sin. O 

Subtract each member of this equation from 1. 

l-cos. a = -<cos^+co f .(i?+^ {(J}) Art 67) 
sm. B sm. C 

' A+B+C B+C-A 

— 2 COS. : ■ COS. — 

= ' ■ 2 g ■ „ ((<?) Art. 70). 

sin. B sin. G. 

Now, let,(l),S= A+ * +C ; (2), ^+^ = 8- A ; 



178 THE ELEMENTS OF SPHERICAL TRIGONOMETRY. 



Substituting values, from (1) and (2), in the last 
equation 

t N .. —2 cos. S cos. (S— A) 

(e) 1— cos. a = ; -^ \ 

sin. B sin. G 

Again, adding 1 to both members of equation (d) 

c 

AC 




^ . cos. ^L+cos. (B— C\ /zx A /ion 

1+cos. a = . ' , v ^ — f (b) Art. 68) 

sm. B sm. C 

A+B-C A+C-B 
2 cos. — — cos. — ■ 



2 



if) 1+cos. a 
(see (3) and (4)). 



sin. i? sin. C 

= 2 cos. (£--ff) c os. (ff-ff) 
sin. J? sin. (7 



or, 



Divide (0) by (f) and extract the square root. 

/ —cos. 8 cos. (S—A) or 
V cob.(S-B)cob.(S-C) j ' 



/I 



-cos. a 



1+cos. « 



r 



tan. i a = '. / ~™»<*«±(*-Sl ((,) Art. 75). 
V cos. (S-JS) cos. (S-C) KKJ ' 

In a similar manner it can be proved 



THEOREMS OF OBLIQUE-ANGLED TRIANGLES. 179 



. z / —cos. S cos. (S—B) 

tan. \b— / - 



cos. (S- C) cos. (S-A) 
x _ / — cos. S cos. (S— C) 

' " T COS. (£-^L) COS. (£-.&) 

149. In order that we may find real values for 
tan.-|#, tan. J&, and tan. \c, the quantity under the 
radical must be positive. This quantity will be posi- 




tive if cos. S is negative (that is, if —cos. S is positive), 
and cos. (S—A), cos. (S—B), cos. (S— C) are all posi- 
tive. 

Now, 2 ^is >180° and <540° (Ch.29, VIII.) ; 

therefore 8 is >90° and <270°; 

consequently cos. S is negative (Art. 88). 

Again, cos. (S—A), cos. (S—B), cos. (#— C) are all 
positive. 

For, let A'B'C be the polar triangle of ABC. 

Then a'=l%Q°-A\ V=180°-B; c'=180°-C 
(Ch. Arts. 69 and 70, VIIL). 

But a! <{V+c f ) (Oh. 25, VIII.), or, 

(180°- A)< 1360°-(B+C)\. 



180 THE ELEMENTS OF SPHERICAL TRIGONOMETRY. 

m . m 2+C-A < lg01 OT<90 o. 

2 2 

But B +°- A -8- A ((2) Art. 148) ; 

2 

therefore S-A< 90°. 

In the same way it can be proved that 

S-B <90° and 8-C <90° ; 

therefore cos. (S— A), cos. (#— i?), cos. (#—(7) are 

all positive (Art. 88). 

im -i ^ 2 sin. (5— J) sin. (5— <?) „ N A , _..,,. 

150. 1— cos. A — > — -* 5- L (M Art. 146) ; 

sin. sin. 

or dividing each member of this equation by 2 and ex- 
tracting the square root, 

/l— cos. A _ 1 sin. (s— h) sin. (s—c) 
V *~2 ~ y sin. J sin. c 

But sin. \A= J — — — - — (Art. 59) ; therefore 



KL 



(a) sin. iA = J ^M^H), 
y sin. J sin. e 

In a similar manner it may be proved 



(J) sin. \B= /&>•(«-«) ™- ('-«). 

y sin. <? sin. & 

( C ) sin. i C= /bJ°- .(«-«) Bin. («-*). 

y sin. a sin. h 

at -i 1 /i 2 sin. 5 sin. (s—a) ,, ,,. A , _,,„, 

Also, 1+cos. J. = — * * ((f) Art. 146). 

sin. 6 sin. c 



THEOREMS OF OBLIQUE-ANGLED TRIANGLES. 181 

Divide each member of this equation by 2 and ex- 
tract the square root. 



/l J- rt A A / 



1+cos. A __ /sin. s sin. (s— a) 



V ' ~2 ~V 



sin. h sin. c 
But cos. \A=J 1+cos '^ (Art. 60) ; therefore 

(J) cos. i A = A°-«™-(— «). 

r sin. 5 sin. c 

In a similar manner it can be proved 
(e) cos. \B= A"- « Bin, (s-b) 



sin. c sin. « 



A 



V sin. a sin. b 

_._ . — 2cOS.#COS. (S— .A),, N A lii(0 v 

151. 1— cos. a = ; — ^-t- a t; '{(?) Art. 148). 

sm. Jj sm. G 



u _ ~ r 



/l— cos. a _ /—cos. ff cos. (S— A) 
V ~2 V sin. ^ sin. (7 



/ 



/ \ • i / — COS.#COS.(,S— ^4) /A , km 

(a) sin. £a = ,/ = — / y (Art. 59). 

V sm. i? sm. G 

In a similar manner it can be proved 

/7A • it / -—cob. S cos. (S—JB) 

(b) sm. J & = V . / — /. 

y sin. (7 sm. ^. 

(«) sin. i o = /-cos- ^ cos. (^-Q^ 
y sin. ^1 sin. i? 



182 THE ELEMENTS OF SPHERICAL TRIGONOMETRY. 

A . ' . 2 cos. (S— B) cos. (S- G) 

Again, 1+cos. a — ^ '- * ' 

sin. B sin. 

((/) Art. 148). 

/ l+cos. a _ / cos. {S— B) cos. {S- C) 
V 2 ~V sin. B sin. G 



V sin. B sin. C 

In a similar manner it can be proved 



(e) cos. i h = /<*»-{S-0)«*.i8-A) . 
V sin. G sin. ^1 

(/) cos. * « = . A* 0?-^) cos- ( gbg). 
V sin. A sin. i> 

152. Gauss's Equations. 

In a spherical triangle 

, . sin. J (J.+^) _ cos, j ( a-5) . 



(5) 



cos. i (7 cos. i <? 

cos. -| (J_+i?) __ cos. -J- (#+&) 



sin. f G cos. -J <? 

M sin, j (J--^) = sin, j Q-5) 
cos. ^ (7 sin. ^ e 

,^ cos.-|(^l— B) _ sin, j Q+£) 
sin. ^ G sin. -J c. 

t, sin. ^4. sin. a ,, N A , ^ OON 
For - 7 — - = -_. «» Art. 138), 
sin. _Z> sm. 6 

sin. ^1+sin. B sin. #+sin. J /jr ., A , _, ^ TTT x 

= (Ch. Art. 10, III.). 

sin. A sin. a 



THEOREMS OF OBLIQUE-ANGLED TRIANGLES. 183 

2 sin, j (A+B) cos, j (A-B) 
sin. A 

= arin.i(a+?)co B .i(a-ft) ((a) Apt 73)< 

sin. & 

By alternation, 

sin. |- (J.+J?) cos. \ (A— B) __ sin. A __ sin.J? 
sin. -| (a+J) cos. ^ (#— 5) sin. a sin. 

((,)Art.l38) = sin -* C ' COS -* 6 ' 
rt. 73) ; the 
(if 



sin. \ c cos. -| 
((a) Art. 73) ; therefore, again by alternation, 
sin, jr (A+B) cos, j (^ -B) 
sin. ^ (7 cos. ^ O 

_ sin. ^ (a-\-b) cos. ^ (a—b) 
sin. -|tf cos. ^<? 

cos, j (j+g = tan, j c 
V ' cos. £ (4--B) tan. £ (a+b) VV ' ; 

tan, j (4+H) = cos, j (a-») ( Arf 
v ; cot. i (7 cos. I (a+5) vv ; ; 

Multiply together equations (1), (2), and (3), mern her 
by memher, cancel like terms in numerator and denomi- 
nator, and extract the square root of the result, and 

sin, j (A+B) = cos- i (fl-h) (equation {a)y 
COS. \ C cos. -J- tf 



Divide equation (1) by equation (4), and 
cos. j(A-B) = sin, j (a+b) . 

sin. -J (7 sin. -J c 

Multiply equation (2) by equation (5). 



184 THE ELEMENTS OF SPHERICAL TRIGONOMETRY. 

, a . cos. -J- (A+B) cos. i (a+b) , , . ,, v 

(6) ■ xTt = i (equation (5);. 

sm. f G 7 cos. f c 

Again, faa-*(^--g) = smj^) Art _ 
cot. i 6 7 sin. ^ (a+&) 

Multiply this last equation by equation (5), and 

(?) *in.HA-B) = sin. * («-ft) (equation {<})) 
cos. |- (7 sin. \ c 



CHAPTER XV. 

SOLUTION OF OBLIQUE-ANGLED TRIANGLES. 

Art. 153. Of the six parts of a spherical triangle 
(the three sides and the three angles), it is necessary to 
know three in order to solve the triangle. 

To solve a right-angled triangle three parts were given, viz., the right 
angle and two other parts. (Arts. 116 and 119.) 

To solve a quadrantal triangle three parts were given, viz., the quad- 
rant and two other parts. (Arts. 129 and 131.) 

154. There are six cases of the solution of oblique- 
angled spherical triangles, which may be classified 
under four heads, as the known parts are, in 

(1) Case I. — The three sides / 

(2) Case II. — The three angles / 

(3) Two sides and an angle, under which head are 
Case III. — Two sides and an included angle, and 
Case IY. — Two sides and an angle opposite one of 

these sides ; 

(4) Two angles and a side, under which head are 
Case V. — Two angles and an included side, and 
Case VI. — Two angles and a side opposite one of 

these angles. 

155. Case I. — The three sides of an oblique-angled 
spherical triangle ieing Tcnown, to solve the triangle. 

Use formulas (a), (i), and (c), Art. 146. 
For check on the work use (c) Art. 138. 



186 THE ELEMENTS OF SPHERICAL TRIGONOMETRY. 



Example. Suppose ABC to be a spherical triangle, of which < 
6=40°, c=76°. 



=50°, 




a = 50 ( 

b = 40" 
c = 76 
166 

2 



a-f 6-t-c 



= = 83° 



2 

s— a = '6'd 
s-b = 43 
«— c = 



*2° 



Tan. 1^1 



_ A / sin. (6 

V, 



s — 6) sin. (s — c) 
s sin. (s — a) 



sin. 43° sin. 7° 



1 sin. 33° 



Tan. i B 



Log. sin. 43° = 9.833783 

Log. sin. 7° = 9.085894 

Ar. co. log. sin. 83° = 0.003249 

Ar. co. log. sin. 33° = 0.263891 

2 | 19.186817 
Log. tan. 21° 24' 38.47"= 9.593*085 
.-. A = 42° 49' 16.94*. 



__ / sin, (s—c ) sin, (s— a) 
f sin 

f si 



sin. s sin. (s—b) 
sin. 7° sin. 33° 



sin. 83° sin. 43° 

Log. sin. 7° = 9.085894 

Log. sin. 33° = 9.736109 

Ar. co. log. sin. 83° = 0.003249 

Ar, co. log. sin. 43° = 0.166217 

2 [ 18.991469 
Log. tan. 17° 23' 14.18'= 9.4957345 
. •. ^=34° 46' 28.36". 



Tan, j C = 4 /™' (*-"> sin - (*-*> 

r SLD 



=/' 



sin. s sin. (s—c) 
sin. 33° sin. 43° 
sin. 83° sin. 7° 
Log. sin. 33° = 9.736109 
Log. sin. 43° = 9.833783 
Ar. co. log. sin. 83° = 0.003249 
Ar. co. log. sin. 7° = 0.914106 
2 J 20.487247 
Log. tan. 60° 17' 18.23" = 10.2436235 
. • . C = 120° 34' 36.46". 



SOLUTION OP OBLIQUE-ANGLED TRIANGLES. 187 

frn _ . sin. A sin. B sin. C 

(Check.) — = — — - = 

sin. a sin. o sui. c 

Log. sin. 42° 49' 16.94" = 9.832327 Log. sin. 34° 46' 28.36" = 9.756140 

Log. sin. 50° = 9.884254 Log. sin. 40° = 9.808067 

1.948073 1.948073 

Log. sin. 120° 34' 36.46" = 9.934977 
Log. sin. 76° = 9.986904 

1.948073 

Find the angles of a triangle when the sides are : 
Example 1. 75°, 100°, 65°. Arts, 68° 8' 54.4", 108° 51' 46.2", 

60° 33' 33.3". 

2. 57°, 83°, 114°. Arts. 49° 3' 20", 63° 22' 18.3", 124° 38' 8.6". 

3. 70°, 140°, 80°. Ans. 41° 22' 18.7", 153° 7' 14.7", 43° 50' 32.1". 

4. 120°, 80°, 60°. 

5. 100°, 69°, 51°. 

156. Case II. — The three cmgles of an oblique-an- 
gled spherical triangle being known, to solve the triangle. 
Use formulas (a), (S), and (c) of Art. 148. 
For check on work use {e) Art. 138. 

Example. Suppose the angles A, i?, and C of a spherical triangle to 
be 130°, 60°, and 74° respectively, required the sides. 



Ta D .4a=4/~ COS ^ C08 - ( ^-^> 



A = 130° 
B = 60° 

C = 74° 


:132° 


'V cos. (S-B) cos. (JS-C) 

/ — cos. 132° cos. 2° 
V cos. 72° cos. 58° 


,_A + B + C __ 264°_ 
2 2 
JS-A = 2° 
S—B— 72° 


Log. (— cos. 132°) = 9.825511 
Log. cos. 2° = 9.999735 
Ar. co. log. cos. 72° = 0.510018 
Ar. co. log. cos. 58° = 0.275790 


S-C= 58° 


2 | 20.611054 



Log. tan. 63° 40' 17.55"=10.305527 
.-. a =127° 20' 35.1". 



188 THE ELEMENTS OF SPHERICAL TRIGONOMETRY. 



Tan. ib=A/- c ^ S cos - ^-^ Tan. 4 e / - cos. ^ cos. (>9 - Q 

r cos. (£-<?) cos. (£-^1) y cos.(&-^)cos.(£-J3) 

_ / — cosTl32° cos. 72° / —cos. 132° cos. 58° 

r cos. 58° cos. 2° ~~ V cos. 2° cos. 72° 

Log. (— cos. 132°) = 9.825511 Log. (— cos. 132°) = 9.825511 

Log. cos. 72° = 9.489982 Log. cos. 58° = 9.724210 

Ar. co. log. cos. 58° = 0.275790 Ar. co. log. cos. 2° = 0.000265 

Ar. co. log. cos. 2° = 0.000265 Ar. co. log. cos. 72° = 0.510018 

2 I 19.591548 2 I 20.060004 



Log. tan. 31° 59' 56.81"= 9.795774 Log. tan. 46° 58' 39.05"=10.030002 
. • . b = 63° 59' 53.62". . • . c = 93° 57' 18.1". 

trrt T N sin. A sin. B sin. C 

( Check.) —. = — — r = — 

sin. a sin. o sin. c 

Log. sin. 130° = 9.884254 Log. sin. 60° = 9.937531 

Log. sin. 127° 20' 35.1" = 9.900377 Log. sin. 63° 59' 53.6" = 9.953654 

1.983877 1.983877 

Log. sin. 74° = 9.982842 
Log. sin. 93° 57' 18.1" = 9.998965 

1.983877 

Find the sides of a triangle when the angles are : 
Example 1. 64°, 75°, 117°. Arts. 68° 6' 36.4", 85° 43' 1.4", 

113° 5' 38.7". 

2. 136°, 72°, 102°. Am. 147° 23' 9.2", 47° 33' 14.4", 130° 37' 46.8". 

3. 125°, 75°, 100°. 

4. 143° 3', 119° 12', 110° 35". Ans. 140° 10' 43J", 111° 34' 17.4". 

85° 48' 47.9". 

5. 129° 5' 20", 105° 8', 142° 12' 40". 

157. Case III. — Two sides and an included angle 
of a spherical triangle being known, to solve the triangle. 

To find the angles : first, find the half sum and the 
half difference of the angles, opposite the given sides, 
by Art. 144 ; then add these quantities for the greater 
angle, and subtract the half difference from the half 
sum for the less angle. 



SOLUTION OF OBLIQUE-ANGLED TRIANGLES. 189 

To find the third side use (a) or (b) of Art. 143, or 
use Art. 138. 

If the third side is found by Art. 143, for check on 
work use (c) Art. 138, otherwise use a formula of Art. 
143. 

Remark. — When -J- (a— b\ and consequently -J ( A— B\ is a very small 
quantity, -J- c is better found by (a) of Art. 143 than by (b) of same article, 
since the cosines of small angles differ from one another less than the 
sines (as will be seen from an inspection of the tables), and the effect of a 
small error in A or B would be greater if sin. -J (A— B) were used rather 
than cos. 4- (A—B). 

On the other hand, if \ (a + 6), and consequently -J (A + B) y is near 
90°, to find i c (b) Art. 143 should be preferred, since near 90° the sines 
differ less than the cosines. 

Example. Suppose ABC to be a spherical triangle, of which the side 
a=76°, 6=55° 10', and the angle (7=125°. 
a— 16° 
b= 55° 10' 



131° 10' n , 

i(a + b) = =65° 35'; 

2 

20° 50' 

i (a-b) = = 10° 25' ; 

2 

i C=62° 30'. 




cos. -J- (a— b) 

tan. i (A + B) = - cot. £ (7; 

cos. ^ (a + b) 

cos. 10° 25' 

= cot. 62° 30' ; 

cos. 65° 35' 

Log. cos. 10° 25' ==9.992783 

Log. cot. 62° 30'=9. / 71647T 

Ar. co. log. cos. 65° 35'=0.383662 

Log. tan. 51° 4' 59.V7"=10.092922 

sin. -A- (a—b) 

tan. i (A-B) = } cot. \ C; 

Sin. $ (a + b) 

sin. 10° 25' 

= cot, 62° 30' ; 

sin. 65° 35' 



190 THE ELEMENTS OF SPHERICAL TRIGONOMETRY 

Log. sin. 10° 25' = 9.257211 

Log. cot. 62° 30' = 9.716477 

Ar. co. log. sin. 65° 35' = 0.040690 

Log. tan. 5° 54' 5.34" = 9.014378 

i (A+B) = 51° 4' 59.77" 
4 (A-B) = 5° 54' 5.34" 

A = 56° 59' 5.11" 
-5=45° 10' 54.43" 

cos. -J (A + B) 

tan. i c = tan. % (a+b) ; 

cos. i (A-B) 

cos. 51° 4' 59.77" 

== tan. 65° 35'. 

cos. 5° 54' 5.34" 

Log. cos. 51° 4' 59.77" = 9.798092 

Log. tan. 65° 35' = 10.342972 

Ar. co. log. cos. 5° 54' 5.34" = 0.002308 



(Cheek.) 



Log. tan. 54° 17' 24|" 10.143372 
. • . c = 108° 34' 49V 
sin. A sin. B sin. O 



sin. a sin. 6 sin c 

Log. sin. 56° 59' 5.11" = 9.923516 
Log. sin. 76° = 9.986904 

1.936612 

Log. sin. 45° 10' 54.43" = 9.850858 Log. sin. 125° = 9.913365 

Log. sin. 55° 10' = 9.914246 Log. sin. 108° 34' 49j" = 9.976753 

1.936612 1.936612 



Solve a spherical triangle when there are given the two sides and the 
included angle : 

Example 1. Sides 112° 30', 60° 15'; angle 35° 40'. 

Ans. Angles 142° 36' 17.7", 34° 47' 58.1"; side 62° 29' 57.4", 

2. Sides 77° 41', 54° 16' ; angle 122° 13'. 

Am. Angles 59° 59' 36.7", 46° 0' 49.2" ; side 107° 21' 2.7", 

3. Sides 140°, 113° 22'; angle 110° 16'. 

Ans. Angles 142° 41' 58.2", 120° 3' 59.9" ; side 84° 17' 40.3" 

4. Sides 105°, 65° ; angle 40°. 



SOLUTION OF OBLIQUE-ANGLED TRIANGLES. 191 

158. Case IV. — Two sides and an angle opposite 
one of these sides being known, to solve the triangle. 

The angle opposite the other given side can be found 
by means of Art. 138. 

The third side may be found by either of the for- 
mulas of Art. 143. Formula (a) should be preferred if 
£ (a—b) is a small quantity, but formula (b) if \ (a-\-b) 
is near 90° (see remark under preceding Art.). 

The third angle may be found by either of the for- 
mulas of Art. 144. Formula (a) is to be preferred if 
-| {a—b) is a small quantity, but formula (b) if i (a-\-b) is 
near 90° (on the principle given in remark of preceding 
Art.). 

As check use Art. 138. 

As the first required part is found from its sine, and 
as the sine of an acute angle and the sine of its supple- 
ment are the same, both in value and in sign, the part 
found may be either less than 90° or greater than 90° 
(Art. 46). 

In some examples, therefore, there may be two tri- 
angles having the given parts, and in such examples we 
shall obtain two solutions, each of which will be correct. 

This will be evident from the following figure. 

Suppose the given parts to be a, b, and the angle A. 

Let the triangle be projected upon the plane of the 
base of the hemisphere, of which the circumference 
contains the side e as a part of it. Let CD be the pro- 
jection of the perpendicular from C (to the arc ABD). 

From C two arcs can be drawn to the circumference 
ABB', each equal to a, one on each side of CD (Art. 
132). Thus the triangles ABC and AB f C each con- 
tain the angle A, the side S, and the side a. 



192 THE ELEMENTS OF SPHERICAL TRIGONOMETRY. 




There will not always be two solutions under this 
case. For instance, if a + J =180°, and A is given, B 
will be known, since ^.+^=180° (Art. 136). 

159. Two sides and an angle opposite one of these 
being given, to determine whether the parts belong to 
one triangle or to two triangles / that is, whether there 
should be one solution or two solutions. 

Considering alone the sum of the given sides, we 
may bring all examples under three heads, as the sum 
of the given sides is : 

1. equal to 180° ; 

2. less than 180° ; 

3. greater than 180°. 

Under all these heads there will be one solution if 
the given sides are equal (Ch. 23, VIII.). 

In what follows, therefore, we shall consider the 
sides unequal. 

1. Let the sum of the given sides equal 180°. 

There will be one solution, since the sum of the 
opposite angles must equal 180° (Art. 136). 



SOLUTION OF OBLIQUE-ANGLED TRIANGLES. 193 

Thus, suppose a+5=180°, and A is given ; then 
A+B=180° ; therefore ^=180°—^. 

2. Let the sum of the given sides be less than 180°. 

Suppose a+b <180°, and A be the given angle : 

(a) Let a be >b ; then there will be one solution. 

For suppose A to be >90°, then B must be <90° 
(Art. 136) ; but suppose A to be < 90°, then B must be 
<90°(Ch. 26, VIIL). 

(i) Let a be < by and A < 90 ; there will be two 
solutions, since B might have two values, one < 90 and 
greater than A, or another >90°, and therefore > A (Ch. 
26, YIII.), and A+B would still be <180° (Art. 136). 

Thus, suppose a-f&<180°, and ^=35°, B might be 50° or 130°, 
as with either value A is <i?, and -4 + i?<180 o . 

When a + 6<180°, and a is <6, and A is >90°, no triangle can be 
formed, for B would also be >90(Ch. 26, VIIL), and A + B would be 
>180°, which is impossible (Art. 136). 

The first and second classes of examples may be illustrated by the 
accompanying figure, which represents the surface of a hemisphere pro- 




jected upon the plane of its base (Art. 133). CAB represents a spherical 
triangle. CD represents the perpendicular from C to the circumference 



194 THE ELEMENTS OF SPHERICAL TRIGONOMETRY. 

of the base of the hemisphere. ACB represents a seinieircumference 
(Ch. Art. 32, VIII.). CE and CE' represent two equal arcs. CB is 
an arc of a great circle, drawn from (7, equal to CB (Art. 132). 

1. a + b=!80°=AC+CB. 

2. (a)a + 6<180 3 ; (a> 6, A (=E)>90°)=AC+ CE. 

(a> b,A(=E) <90°)=^<7+ CE. 
(&)a + 6<180°, a<b and A<90°=EC+ CA, ovE'C+CA. 

3. Let the sum of the given sides be greater than 
180°. 

(a) Suppose a+5>180° and a is <b. 

Now, whether A is < 90° or >90°, B must be >90°, 
and so" there will be but one solution ; for 

If A is <90°, A+B >180° (Art. 136), and there- 
fore B must be >90°, and can have but one value, and 

If A is >90°, B is >A (Ch. 26, VIIL), and there- 
fore B must be >90°, and can have but one value. 

(i) If, now, a is >b and A >90°, i? may be either 
< 90° or >90°, for, with either value of B, A+B might 
be >180°, and A>B (Ch. 28, VIIL), so that in this 
case there would be two solutions. 

Thus, suppose a + b>180° 1 a>b, and ^4—110° ; B might be 15° or 
105°. With either value of B, A is >B, A + B> 180°. 

When « + &> 180° and a is >6, and ^4<90°, no triangle is formed; 
for B must be <A, i. e., <90°, and A + B would be <180°, which would 
be impossible. 

The last case, (3), when a + b> 180°, may also be illustrated by a figure. 

Let the hemisphere be projected upon the plane of its base (Art. 133). 

CA, CA! , CE, and CB are all arcs of great circles drawn from C. 
CA and CA! are equal. CD' is the longer perpendicular from C upon 
the circumference of the base of the hemisphere. 

(a) a + b> 180° ( a <b,A<W°)=CB+CA ) 

(a<b, A>90°)=CB+ CA' S { '' 

(b) a + b> 180° (a>b, A>90°)=CE+CA or CE+CA'. 

Therefore, when two sides of a spherical triangle 



SOLUTION OF OBLIQUE-ANGLED TRIANGLES. 195 




and an angle opposite one of them are given (consider- 
ing only possible cases), 

If the sum of the given sides is less than 180°, and 
if the side opposite the given angle is less than the other 
given side ((b) 2), or, 

If the sum of the given sides is greater than 180°, 
and if the side opposite the given angle is greater than 
the other given side ((b) 3) : 

There will be two solutions ; in all other cases 
there will be one solution. 




Example. Suppose «=62° 40', b=lS° 13', and .4=44° 18'. 
As a + b=135° 55'<180°, and as A is <90°, there are two solu- 
tions. (See rule.) 



Sin.B = 



si n, b sin. A _ sin. 1S° 13' sin. 44° 18' 
sin. a sin. 62° 42' 



196 THE ELEMENTS OF SPHERICAL TRIGONOMETRY. 

Log. sin. 73° 13' =9.981095 

Log. sin. 44 c 18' =9.844114 | (b + a)=61° 57' 30' 

Ar. co. log, sin. 62= 42" =0.051285 i (b-a)= 5° 15' 30" 

Log. sin. 48"' 45 20"=9.876494 
B = 48* 4^ 20 i(B+A)=46 € 33' 10' i(JT— JL)= 2° 15' 10" 

B =131° 11 40" J (5 -f^)=87° 44' 50" i (B -^)=43° 26' 50" 

Solving triangle AB C, 

™ - cos. ~ \B'+A) w 7X cos.46°33'10" 

Tan. * c= =—- A tan.| a + 6 = — -^ tan. 67° 57' 30" 

cos. | ^ -^) cos. 2° 15' 10" 

Log. cos. 46° 33' 10' = 9.837390 

Log. tan. 67° 57' 30" =10.392682 

At. co. l og, cos. 2 = 15' 10" = 0.000336 

Log. tan. 59° 31' 55.62'= 10.230408 

c=AB f =119° 3' 51.25" 

" , r , cos. \ (6 + a) . . W1 cos. 67° 57' 30" a , „ 

Cot. i C= --^- tan. i {A-rB)= tan. 46° 33' 10" 

cos. J (b-a) - cos. 5° 15' 30" 

Log. cos. 67° 57' 30" = 9.574356 

Log. tan. 46° 33' 10" =10.023551 

Ar. co. log, cos. 5° 15' 30" = 0.001832 

Log. cot. 68° 18' 14.26'= 9.599739 

C=ACB , =ISQ° 36' 28.52" 

Solving triangle ABC, 

sin. l(B+A) n , sin. 87° 44' 50" a , „ 

Tan. I c= -— xajiA'b-a = tan. 5° 15' 30" 

sin. I (B-A) sin. 43° 26' 50" 

Log. sin. 87° 44' 50" =9.999664 

Log. tan. 5 = 15' 30* =8.963947 

Ar. co. log. sin. 43° 26' 50" = 0.162610 

Log. tan. 7° 37' 0.6"=9.126221 

c=AB=Vd z 14' 1.25" 

Cot. i fe ' 8 * *<» + «- ' tan. , (B-A) ^*™'™'. **■«• 26' 50" 
sin. i (b—a) " sin. 5° 15 30" 

Log. sin. 67° 57' 30" = 9.967038 

Log. tan. 4S : 26' 50' = 9.976449 

Ar. co. log. sin. 5 : 15' 30" = 1.037884 



Log. cot. 5"" 57' 32.79' = 10.981371 
C=ACB=IV 55' 5.58' 



SOLUTION OF OBLIQUE-ANGLED TRIANGLES. 197 

sin. B sin. A CB' sin. A CB 

(Check.) - — - = —. — = — — 

sin. 6 sin. AB sin. AB 

Log. sin. 48° 48' 20"=9.876494 Log. sin. 136° 36' 28.52"=9.836948 

Log. sin. 73° 13' = 9.981095 Log. sin. 119° 3' 51.25"= 9.941549 

1.895399 1.895399 

Log. sin. 11° 55' 5.58"=9,314952 

Log. sin. 15° 14' 1.25"= 9.419553 

1.895399 
Solve a triangle when there are given : 

Example 1. Sides, 62° 14', 50° 3' ; angle opposite latter, 35° 33'. 
Ans. Angles, 131° 25' 9.6", 42° 9' 5.8" ; side, 98° 36' 12.1" ; 
or 11° 45' 13.8", 137° 50' 54.2" ; or 15° 34' 49.7". 

2. Sides, 135° 10', 115°; angle opposite first, 143°. 

Ans. Angles, 50° 40' 44J", 23° 13' 13.6" ; side, 27° 30' 35.2" ; 
or 129° 19' 15f", 121° 27' or 91° 55' 50.5". 

3. Sides, 137° 2', 145°; angle opposite first, 151°. 

Ans. Angles, 155° 55' 16.8", 137° 28' 16.4"; side, 71° 51' 40". 

4. Sides, 132° 10', 63° 20' ; angle opposite latter, 63° 20'. 

160. Case Y. — Two angles and an included side of 
a spherical triangle being known , to solve the triangle. 

To find the half sum and half difference of the sides, 
use formulas of Art. 143. 

The sum of these differences will be the greater 
side ; the difference of these differences will be the less 
side. 

The third angle may be found by Art. 144. 

Use formula (a) if |- (a—b) is very small; but for- 
mula (b) if ^ (a+b) is near 90°. 

As check on work use (c) Art. 138. 

Thus, suppose the parts given are A, B, and c. 



m , , 7X cos. \ (A—B) 

Tan. i (a + b) = * tan. i c 

cos. \ (A + B) 

. , , .. sin. i (A—B) 4 , 
tan. i (a-b) = - — v ' tan. i c 

sm. -J- (A + B) 



will give a and b. 



198 THE ELEMENTS OF SPHERICAL TRIGONOMETRY. 

cot.iC = COS -tS a + ; ) tan. i (^ + ^),or 
cos. \ (a—b) 

„ „ sin, 4- (a + b) , , _ 

cot i C= — -77 ^ tan - * (^-^)» T " m g iv e C. 

sm. J (a—b) 

Solve a triangle when there are given : 

Example 1. Two angles, 33° 16', 12° 14'; the included side, 49° 14'. 
Ans. Sides, 38° 14' 16.8", 13° 49' 57.2"; angle, 137° 50' 14.8". 

2. Two angles, 140° 10', 110° 2'; the included side, 125° 4'. 
Ans. Sides, 138° 37' 58.8", 75° 45' 58.8"; angle, 127° 30' 12.8". 

3. Two angles, 78°, 56° ; the included side, 59°. 

Ans. Sides, 61° 33' 39.8", 48° 10' 59.3" ; angle, 72° 27' 31.8". 

4. Two angles, 150°, 110°; the included side, 130°. 

5. Two angles, 40°, 82 ° ; the included side, 75°. 

161. Case VL — Two angles and a side opposite one 
of the given angles of an oblique-angled spherical tri- 
angle being known, to solve the triangle. 

The side opposite the other given angle may be 
found by Art. 138. 

The third side may then be found by Art. 143. 

Use formula (a) if one half the difference of the 
given angles is a very small quantity, but use (b) if one 
half the sum of the given angles is near 90°. (See re- 
mark under Art. 157.) 

The third angle may be found by Art. 144. 

Use formula (a) if one half the difference of the 
sides opposite the given angles is a very small quantity, 
but use (b) if one half the sum of the same sides is 
near 90°. 

As check on work use Art. 138. 

As the first side is found by means of its sine, and 
as the sine of an angle and the sine of its supplement 
are the same (Art. 46), there may be two values given 



SOLUTION OF OBLIQUE-ANGLED TRIANGLES. 199 

to the side, either of which will satisfy the conditions of 
the problem. Consequently there may be two triangles 
with the given parts, and in such a case two solutions 
are possible, both of which will be correct. This will 
be evident from the accompanying figure, in which the 




surface of a hemisphere is represented as projected 
upon the plane of its base (Art. 133). 

Suppose the angles CAB and CBA and the side 
CB are the given parts. Let the arc CA be produced 
to meet the plane of the base of the hemisphere again 
at A\ so that ACA f is a semicircumference (Ch. Art. 
32, VJLJLL). Suppose the given side CB is not equal to 
90° ; then C is not the pole of the base (Ch. Art, 37, 
VIIL). Since C is not the pole of the base, from C 
two arcs CB and CB' can be drawn to the base equal 
to one another (Art. 132). The angles CBB' and 
CB'B are equal (Ch. 23, VIIL) ; therefore their sup- 
plements CB A and CB A' are equal. Also A is equal 
to A' (Ch. 16, VIIL). 



200 THE ELEMENTS OF SPHERICAL TRIGONOMETRY. 

Thus we have two triangles CAB and CA'B', 
which have two angles in one equal respectively to two 
angles in the other, and the side opposite one of the 
equal angles the same in both. The side opposite the 
other given angle of one is the supplement of the side 
opposite the equal given angle of the other. Thus, CA 
opposite CBA is the supplement of CA\ which is 
opposite CB'A! (the equal of CBA). 

162. Two angles and a side opposite one of these 
being given, to determine whether these parts "belong to 
one or to two triangles / that is, whether we shall have 
one solution or two solutions from the given parts. 

From the given parts we may find the two sides 
and an angle opposite one of these sides in the polar tri- 
angle (Ch. Arts. 69 and 70, VIII.). There will be as 
many solutions for the triangle whose parts are given 
as for the polar triangle. But we have already consid- 
ered the case of the triangle of which two sides and an 
angle opposite one of these sides are given (Art. 158). 
By passing to the polar triangle we can then find (under 
Case IY. and Art. 159) how many solutions it has, and 
thus determine how many solutions the triangle under 
this case has. 

We can also ascertain this independently by refer- 
ring to Ch. 26, VIII., and to Art. 137 of this book. 

Thus, suppose the given parts are the angles A and 
B, and the side a. 

The sum of the angles may be, 1. 180° ; 2. <180° ; 
or, 3. >180°. 

Under these three heads, if A—B, a=b (Ch. 24, 
VIII.), and there is one solution. 

In the following cases we shall suppose A and B 
unequal. 



SOLUTION OF OBLIQUE-ANGLED TRIANGLES. 201 

1. IfA+B=l$0°, a+b also equals 180° (Art. 137), 
and there can be but one solution. 

2. Let A+B he <180°; and 

(a) Let A be >B ; there will be one solution; for 
if a is <90°, b must be <a (Ch. 26, VIII.) ; and if 
a is >90°, b must be <90°, since a+b <180° (Art. 
137). 

(b) Let A be <B, and a<90° ; then there will be 
two solutions, for as A is < J?, a is < b ; but 5 may have 
two values, one <90° but >&, and the other >90°, 
such that a+b <180°. 

Thus, suppose ^ + ^<180°, and A<B and a=40° ; b might be 60° 
or 120°, for with either value a is <6 and a-f-S<180°. 

The case (under this head) in which A is <B and a > 90°, is impos- 
sible, for if A is <i?, a is <5, . ■ . 6 is also > 90°, that is, a + 6> 180°, which 
is impossible (Art. 137). 

3. Let A+B he >180°, and 

(a) Let J. be < J? ? then there will be one solution ; for 
if A <B, a<b (Oh. 26, VIII.) ; therefore if a<90°, b 
must be >90°, so that a+b should be >180° (Art. 
137); or again, if <z>90°, b must be >90°, because a 
is < b as by hypothesis A is <B (Ch. 26, VIIL). 

(b) Let A be >B, and a be >90° ; then there will be 
two solutions, for as A is >B, a is >5 (Ch. 26, VIIL) ; 
but b may have two values, one < 90°, and therefore < a, 
or one >90°, but still <a, and a+b >180° (Art. 137). 

Thus, suppose A + B>180° 1 and a to be 120°, and A>B, then b 
might be 75° or 105° ; with either value a is >6, and a + 5>180°. 

The case (under this head) in which A is >i?, and a is <90°, is im- 
possible ; for as A is >B,a must also be >6 ; therefore b must be <90°, 
that is, a+b is <180°, which is impossible (Art. 137). 

Therefore (considering only possible cases), when two 



202 THE ELEMENTS OF SPHERICAL TRIGONOMETRY. 

angles of a spherical triangle and a side opposite one of 
them are given, 

If the sum of the given angles is less than 180°, and 
the angle apposite the given side is less than the other 
given angle ((b) 2), or, if the sum of the given angles is 
greater than 180°, and the angle opposite the given side 
is greater thorn, the other given angle ((b) 3), there will 
be two solutions ; in all other cases only one solution. 

The two angles of a triangle are 125° and 110°, and the side opposite 
the first angle is 133°. It is required to solve the triangle. 
Let A = 125°, B = 110°, a = 133°. 
A + B >180°, and A is >B; therefore there are two solutions. 




« x sin - 5 • sin. 110° 

Sin. b = sin. a = — — — — sin. 133 

sin. A sin. 12o 

Log. sin. 110°= 9.972986 

Log. sin. 133° = 9.864m 

Ar. co. log. sin. 125° = 0.086635 

Log. sin. 57° 1' 54.62" = 9.923748 
b = CA = 57° 1' 54.62" i (A + B) = 117° 30' 

$'=<7^'=122°58' 5.38" \(A—B)= 7° 30' 

To solve ABC 
£(a + 6) = 95° 0' 57.31" 
i (a-b) = 37° 59' 2.69" 



SOLUTION OF OBLIQUE-ANGLED TRIANGLES. 203 

coaA(A + B) . _. cos. 117° 30' ftl . OA ,. l . (i<l1l 

Tan. *<? = 7V y tan. i(a + 6) = tan. 95° 0' 57.31" 

cos.i(A-B) V ' cos. 7° 30' 

Log. cos. 117° 30' = 9.664408 

Log. tan. 95° 0' 57.31" = 11.056661 

Ar. co. log. cos. 7° 30' = 0.003731 



Log. tan. 79° 19' 39.48" = 10.724798 
c = ^^=158°39' 18.96". 

Y cos. i (a— 6) cos. 37° 59' 2.69" 

Log. cos. 95° 0' 57.31" = 8.941673 
Log. tan. 117° 30' = 10.283523 
Ar. co. log. cos. 37° 59' 2.69" = 0.103374 

Log. cot. 77° 58' 14.08" = 9.328570 
A (7.5=155° 56' 28.16". 
To solve A' B C 
i(a + b') = 12l° 59' 2.69" 
$la-b')= 5° 0' 57.31" 

Tan. i <f = C ° S - m ° 30 ' tan. 127° 59' 2.69" 
cos. 7° 30' 

Log. cos. 117° 30' = 9.664406 

Log. tan. 127° 59' 2.69" = 10.107439 

Ar. co. log. cos. .7° 30' = 0.003731 



Log. tan. 30° 48' 50.62^" = 9.775576 
c' = A'B' = 61° 37' 41.25". 

„ , - cos. 127° 59' 2.69" 

Cot i C = tan. 117° 30' 

cos. 5° 0' 57.31" 

Log. cos. 127° 59' 2.69" = 9.789187 

Log. tan. 117° 30' = 10.283523 

Ar. co. log. cos. 5° 0' 57.31" = 0.001667 



Log. cot. 40° 7' 3.26" = 10.074377 
A 1 C B = 80° 14' 6.52". 

sin. B sin. C sin. O 

( Check.) - — - = = 

sin. b sin. c sin. c 

Log. sin. 110° = 9.972986 Log. sin. 155° 56' 28.16"= 9.610314 

Log. sin. 57° V 54.62" = 9.923748 Log. sin. 158° 39' 18.96"= 9.561076 

0.049238 0.049238 



204 THE ELEMENTS OF SPHERICAL TRIGONOMETRY, 

Log. sin. 80° 14' 6.52" = 9.993662 
Log. sin. 61° 37' 41.25" = 9.944424 

0.049238 

Solve a spherical triangle when there are given : 
Example 1. Two angles, 55° 2', 68° 10' ; the side opposite the first 
angle, 48° 42'. 

Ans. Sides, 58° 4' 58.6", 65° 26' 30.6" ; angle, 84° 5' 12.4"; 
or 121° 55' 1.4", 160° 12' 23.8" ; or 158° 15' 51.1". 

2. Two angles, 150° 22', 124° 12' ; the side opposite the first angle 
149° 20'. 

Ans. Sides, 58° 33' 28£", 143° 33' 22" ; angle, 144° 50' 21.8" ; 
or 121° 26' 31|", 73° 18' 35.6" ; or 111° 47' 4.9". 

3. Two angles, 125°, 130°; the side opposite the first angle, 114°. 

Ans. Sides, 121° 18' 56.2", 98° 36' 17.2"; angle, 117° 33' 11.7". 

4. Two angles, 12°, 25° ; the side opposite the second angle, 60°. 

SOLUTION OF OBLIQUE-ANGLED TRIANGLES BY MEAKTS OF 
RIGHT-ANGLED TRIANGLES. 

163. The preceding cases can all be solved by divid- 
ing the given triangle into two right-angled triangles, 
by drawing a great circle arc from one of the angles to 
the opposite side, or to the opposite side produced, and 
by then solving the two right-angled triangles. 

Care should be taken, in drawing the perpendicular, 
to make one of the right-angled triangles formed con- 
tain two of the three given parts. (Art. 116.) 

Case I. — The three sides heing given. 

From the angle opposite the longest side draw a per- 
pendicular to that side. The side will be thus divided 
internally into two segments. The half difference of 
the two segments can be found by formula (a) of Art. 
142. This half difference added to the half side will 
give the greater segment ; and subtracted from the half 
side will give the less segment. We shall then have, 



SOLUTION OF OBLIQUE-ANGLED TRIANGLES. 



205 



in each right-angled triangle, the hypotenuse and a side 
given to find the other parts. (Art. 120.) 

Case II. — The three angles being given. 

Find the sides of the polar triangle (Oh. Arts. 69 
and 70, VIII.) ; solve that by the method of the pre- 
ceding case ; the supplements of the angles found will 
be the required sides of the given triangle. 

Case III. — Two sides and an included angle being 
given. 

Suppose the given parts to be 
#, 6, and C. Draw the perpen- 
dicular AD. Then in the tri- 
angle A CD we have the hypote- 
nuse and an angle, (7, to find CD 
(Art. 121). BD=a-CD. Then 
c can be found by 2, Art. 140. A and B can be found 
by Art. 138. The three last parts can also be found 
by finding all the parts of the two triangles A CD \ BCD. 

Case IV. — Two sides and an angle opposite one of 
these sides being given. 

Suppose the given parts to be a, 2>, and A. Draw 
the perpendicular from C upon c or c produced. 

C 





The figure represents one of the examples in which there are two 
solutions (see rule at the end of Art. 159). We shall only give one solu- 
tion, as the method is the same in both solutions. 

To solve the triangle A CD we have given the hy- 



203 THE ELEMENTS OF SPHERICAL TRIGONOMETRY. 



potenuse b and the angle A (Art. 121). BD can be 
found by 2, Art, 110. Then to solve the triangle BCD 
we have given the hypotenuse a and the side BD 
(Art. 120). 

Case Y. — Two angles and an included side being 
given. 

Suppose the given parts to be A, B, and c. Draw 
an arc from B perpendicular to b. 
To solve the triangle ABD we 
have the hypotenuse c and the an- 
gle A (Art. 121). This will enable 
us to find BD and the angle ABD. 
If BD falls within the triangle, 
if without the triangle, CBD— 
ABD—B. Then to solve CBD we have given the 
side BD and the angle CBD (Art. 123). 

Case VL — Two angles and a side opposite one of 
these being given. 

(See rule at end of Art. 162.) 

Suppose the given parts to be A, B, and a. Draw 
the perpendicular from C upon o (or upon c produced). 




CBD=B-ABD: 




To solve the triangle CBD we have given the hy- 
potenuse a and the adjacent angle (Art. 121). Having 
found BD, AD can be found by 1, Art. 140. AC can 
then be found by 2, 140. Or, by solving the triangle 
CBD we can find CD. Then to solve the triangle 



SOLUTION OF OBLIQUE-ANGLED TRIANGLES. 207 

ACD we have given a side, CD, and the opposite an- 
gle A (Art. 124). C will be the sum of A CD and BCD 
and c will be the sum of AD and i?Z>, if the perpen- 
dicular falls within the triangle ; bat C will be the 
difference of ^4 (7J9 and B 1 CD, and o will be the differ- 
ence of JLJ9 andZ>i?', if the perpendicular falls without 
the triangle. 

164. It is well to solve oblique-angled triangles by 
right-angled triangles when there is any difficulty in 
arriving at a true solution by means of formulas ; for 
instance, when in Case I. s is very near 180°, or s— a, 
s—h,ovs — cis very near 0°; when in Case II. S is very 
near 90° or 270°, or S—A, S-B, or S- C is very near 
90° ; when in Cases III. and IV. \ (a-\-h) is very near 
90°, or \ (a— V) is very near 0° ; when in Cases Y. and 
VI. i (A+B) is very near 90°, or \ (A — B) is very 
near 0°. 

It is sometimes advantageous to combine the two 
methods so as to check the work. 



THE END. 



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& 



THE ELEMENTS 



OF 



SPHERICAL TRIGONOMETRY, 



BT 



EUGENE L. RICHARDS, B. A., 

ASSISTANT PROFESSOR OF MATHEMATICS IN YALE COLLEGE. 



NEW YORK: 
D APPLETON AND COMPANY, 

549 AND 551 BROADWAY. 

1879, 



I 



■ 



